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Exercise 4(A)
1. Find the square of:
(i) 2a + b
(ii) 3a + 7b
(iii) 3a – 4b
(iv) 3a/2b – 2b/3aÂ
Solution:
Using the identities,
(a + b)2 = a2 + b2 + 2ab and
(a – b)2 = a2 + b2 – 2ab
(i) (2a + b)2 = (2a)2 + b2 + 2(2a)(b)
= 4a2 + b2 + 4ab
(ii) (3a + 7b)2 = (3a)2 + (7b)2 + 2(3a)(7b)
= 9a2 + 49b2 + 42ab
(iii) (3a – 4b)2 = (3a)2 + (4b)2 – 2(3a)(4b)
= 9a2 + 16b2 – 24ab
(iv) (3a/2b – 2b/3a)2 = (3a/2b)2 + (2b/3a)2 – 2(3a/2b)(2b/3a)
= 9a2/4b2 + 4b2/9a2 – 2
2. Use identities to evaluate:
(i) (101)2
(ii) (502)2
(iii) (97)2
(iv) (998)2
Solution:
Using the identities,
(a + b)2 = a2 + b2 + 2ab and
(a – b)2 = a2 + b2 – 2ab
(i) (101)2 = (100 + 1)2
= 1002 + 12 + 2×100×1
= 10000 + 1 + 200
= 10201
(ii) (502)2 = (500 + 2)2
= 5002 + 22 + 2×500×2
= 250000 + 4 + 2000
= 252004
 (iii) (97)2 = (100 – 3)2
= 1002 + 32 – 2×100×3
= 10000 + 9 – 600
= 9409
(iv) (998)2 = (1000 – 2)2
= 10002 + 22 – 2×1000×2
= 100000 + 4 – 4000
= 996004
3. Evalute:
(i) (7x/8 + 4y/5)2
(ii) (2x/7 – 7y/4)2
Solution:
Using the identities,
(a + b)2 = a2 + b2 + 2ab and
(a – b)2 = a2 + b2 – 2ab
(i) We have, (7x/8 + 4y/5)2
= (7x/8)2 + (4y/5)2 + 2 × (7x/8) × (4y/5)
= 49x2/64 + 16y2/25 + 7xy/5
(ii) We have, (2x/7 – 7y/4)2
= (2x/7)2 + (7y/4)2 – 2 × (2x/7) × (7y/4)
= 4x2/49 + 49y2/16 – xy
4. Evaluate:
(i) (a/2b + 2b/a)2 – (a/2b – 2b/a)2 – 4
(ii) (4a + 3b)2 – (4a – 3b)2 + 48ab
Solution:
Using the identities,
(a + b)2 = a2 + b2 + 2ab and
(a – b)2 = a2 + b2 – 2ab
(i) Given expression, 
On expanding the term using the identity, we have
(a/2b + 2b/a)2 = (a/2b)2 + (2b/a)2 + 2 × (a/2b) × (2b/a)
= a2/4b + 4b2/a2 + 2
Next, expanding the second term using the identity, we have
(a/2b – 2b/a)2 = (a/2b)2 + (2b/a)2 – 2 × (a/2b) × (2b/a)
= a2/4b + 4b2/a2 – 2
Now, using these results in the given expression
(a/2b + 2b/a)2 – (a/2b – 2b/a)2 – 4 = (a2/4b + 4b2/a2 + 2) – (a2/4b + 4b2/a2 – 2) – 4
= a2/4b + 4b2/a2 + 2 – a2/4b – 4b2/a2 + 2 – 4
= 0
Â
(ii) Given expression, (4a + 3b)2 – (4a – 3b)2 + 48ab
On expanding the term using the identity, we have
(4a + 3b)2 = (4a)2 + (3b)2 + 2 × (4a) × (3b)
= 16a2 + 9b2 + 24ab
Next, expanding the second term using the identity, we have
(4a – 3b)2 = (4a)2 + (3b)2 – 2 × (4a) × (3b)
= 16a2 + 9b2 – 24ab
Now, using these results in the given expression
(4a + 3b)2 – (4a – 3b)2 + 48ab = (16a2 + 9b2 + 24ab) – (16a2 + 9b2 – 24ab) + 48ab
= 16a2 + 9b2 + 24ab – 16a2 – 9b2 + 24ab + 48ab
= 96ab
5. If a + b = 7 and ab = 10; find a – b.
Solution:
Using the identities,
(a + b)2 = a2 + b2 + 2ab and
(a – b)2 = a2 + b2 – 2ab
Rewriting the above equation as
(a – b)2 = a2 + b2 + 2ab – 4ab
= (a + b)2 – 4ab … (i)
We have,
a + b = 7 and ab = 10
So, using these in equation (i), we get
(a – b)2 = (7)2 – (4 × 10)
= 49 – 40
= 9
Then,
(a – b) = √9
= ±3
6. If a – b = 7 and ab = 18; find a + b.
Solution:
Using the identities,
(a – b)2 = a2 + b2 – 2ab and
(a + b)2 = a2 + b2 + 2ab
Rewriting the above equation as
(a + b)2 = a2 + b2 – 2ab + 4ab
= (a – b)2 + 4ab … (i)
We have,
a – b = 7 and ab = 18
So, using these in equation (i), we get
(a + b)2 = (7)2 + (4 × 18)
= 49 + 72
= 121
Then,
(a + b) = √121
= ±11
7. If x + y =Â 7/2 and xy =Â 5/2; find:
(i) x – yÂ
(ii) x2 – y2
Solution:
Using the identities,
(x + y)2 = x2 + y2 + 2xy and
(x – y)2 = x2 + y2 – 2xy
Rewriting the above equation as
(x – b)2 = x2 + y2 + 2xy – 4xy
= (x + y)2 – 4xy … (1)
We have,
x + y = 7/2 and xy = 5/2
So, using these in equation (1), we get
(x – y)2 = (7/2)2 – (4 × 5/2)
= 49/4 – 10
= 9/4
Then,
(x – y) = √(9/4)
= ± 3/2 … (2)
Â
(ii) We know that,
x2 – y2 = (x + y) (x – y)
Substituting values in RHS using given and (2), we get
x2 – y2 = (7/2) (± 3/2)
= ± 21/4
8. If a – b = 0.9 and ab = 0.36; find:
(i) a + b
(ii) a2 – b2
Solution:
Using the identities,
(a – b)2 = a2 + b2 – 2ab and
(a + b)2 = a2 + b2 + 2ab
Rewriting the above equation as
(a + b)2 = a2 + b2 – 2ab + 4ab
= (a – b)2 + 4ab … (1)
We have,
a – b = 0.9 and ab = 0.36
So, using these in equation (1), we get
(a + b)2 = (0.9)2 + (4 × 0.36)
= 0.81 + 1.44
= 2.25
Then,
(a + b) = √2.25
= ±1.5 … (2)
Â
(ii) We know that,
a2 – b2 = (a + b) (a – b)
Substituting values in RHS using given and (2), we get
a2 – b2 = (±1.5) (0.9)
= ± 1.35
9. If a – b = 4 and a + b = 6; find
(i) a2Â + b2
(ii) ab
Solution:
Given, a – b = 4 and a + b = 6
We know that,
(a – b)2 = a2 + b2 – 2ab and
(a + b)2 = a2 + b2 + 2ab
Now, rewriting the above equation as
(a + b)2 = a2 + b2 – 2ab + 4ab
⇒ (a + b)2 = (a – b)2 + 4ab
Substituting the values in the above equation, we get
(6)2 = (4)2 + 4ab
36 = 16 + 4ab
4ab = 36 – 16
4ab = 20
ab = 20/4
(ii) Thus, ab = 5
Now, in the identity: (a + b)2 = a2 + b2 + 2ab
(a + b)2 = (a2 + b2) + 2ab
Let’s substitute the values of the known terms,
(6)2 = (a2 + b2) + 2 × (5)
36 = (a2 + b2) + 10
a2 + b2 = 36 – 10
(i) Thus, a2 + b2 = 26
10. If a + 1/a = 6 and a ≠0 find:
(i) a – 1/a
(ii) a2 – 1/a2
Solution:
Using the identities,
(a – b)2 = a2 + b2 – 2ab and
(a + b)2 = a2 + b2 + 2ab
(i) Now,
(a + 1/a)2 = a2 + 1/a2 + (2 × a × 1/a)
= a2 + 1/a2 + 2 … (1)
Substituting the value of (a + 1/a) in the equation (1), we get
62 = a2 + 1/a2 + 2
36 = a2 + 1/a2 + 2
a2 + 1/a2 = 36 – 2 = 34 … (2)
Similarly,
(a – 1/a)2 = a2 + 1/a2 – (2 × a × 1/a)
= (a2 + 1/a2) – 2
= 34 – 2 … [From (2)]
= 32
⇒ (a – 1/a)2 = 32
a – 1/a = ±√32
= ±4√2 … (3)
Thus, a – 1/a = ±4√2
(ii) We know that,
a2 – 1/a2 = (a – 1/a) (a + 1/a)
Using the given and (3) in the above equation,
a2 – 1/a2 = (±4√2) (6)
= ±24√2
Thus, a2 – 1/a2 = ±24√2
11. If a – 1/a = 8 and a ≠0, find:
(i) a + 1/a
(ii) a2 – 1/a2
Solution:
Using the identities,
(a – b)2 = a2 + b2 – 2ab and
(a + b)2 = a2 + b2 + 2ab
(i) Now,
(a – 1/a)2 = a2 + 1/a2 – (2 × a × 1/a)
= a2 + 1/a2 – 2 … (1)
Substituting the value of (a + 1/a) in the equation (1), we get
82 = a2 + 1/a2 – 2
64 = a2 + 1/a2 – 2
a2 + 1/a2 = 64 + 2 = 66 … (2)
Similarly,
(a + 1/a)2 = a2 + 1/a2 + (2 × a × 1/a)
= (a2 + 1/a2) + 2
= 66 + 2 … [From (2)]
= 68
⇒ (a + 1/a)2 = 68
a + 1/a = √68
= ±2√17 … (3)
Thus, a + 1/a = ±2√17
(ii) We know that,
a2 – 1/a2 = (a – 1/a) (a + 1/a)
Using the given and (3) in the above equation,
a2 – 1/a2 = (8) (±2√17)
= ±16√17
Thus, a2 – 1/a2 = ±16√17
Â
12. If a2 – 3a + 1 = 0, and a ≠0; find:
(i)Â a + 1/a
(ii)Â a2 + 1/a2
Solution:
(i) Given equation,
a2 – 3a + 1 = 0
a2Â + 1 = 3a
(a2 + 1)/a = 3
⇒ a + 1/a = 3 … (1)
(ii) We know that,
(a + b)2 = a2 + b2 + 2ab
Now,
(a + 1/a)2 = a2 + 1/a2 + 2(a)(1/a)
= a2 + 1/a2 + 2
Using (1) in the above equation, we get
(3)2 = a2 + 1/a2 + 2
9 = a2 + 1/a2 + 2
a2 + 1/a2 = 9 – 2
Thus, a2 + 1/a2 = 7
13. If a2 – 5a – 1 = 0 and a ≠0; find:
(i) a – 1/a
(ii)Â a + 1/a
(iii) a2 – 1/a2
Solution:
(i) Given, a2 – 5a – 1 = 0
Rewriting the equation, we get
a2 – 1 = 5a
(a2 – 1)/a = 5
Hence, a – 1/a = 5 … (1)
(ii) We know that,
(a + 1/a)2 = a2 + 1/a2 + 2
Manipulating the above as,
(a + 1/a)2 = a2 + 1/a2 – 2 + 4
(a + 1/a)2 = (a – 1/a)2 + 4
Now, using (1) in the above
(a + 1/a)2 = (5)2 + 4
(a + 1/a)2 = 25 + 4 = 29
Hence, a + 1/a = ±√29 … (2)
(iii) We know that,
a2 – 1/a2 = (a + 1/a) (a – 1/a)
Now, using (1) and (2) in the above equation, we get
a2 – 1/a2 = (5) x (±√29)
Hence, a2 – 1/a2 = ±5√29
14. If 3a + 4b = 16 and ab = 4; find the value of 9a2Â + 16b2.
Solution:
Given, 3a + 4b = 16 and ab = 4
Required to find: value of 9a2Â + 16b2
We know that,
(a + b)2 = a2 + b2 + 2ab
Now, the square of (3a + 4b) will be
(3a + 4b)2 = (3a)2 + (4b)2 + 2 × (3a) × (4b)
= 9a2 + 4b2 + 24ab
And, given 3a + 4b = 16
So, by squaring on both the sides
(3a + 4b)2 = 162
9a2 + 4b2 + 24ab = 256
9a2 + 4b2 + 24(4) = 256 [Given ab = 4]
9a2 + 4b2 = 256 – 96
⇒ 9a2 + 4b2 = 160
15. The number a is 2 more than the number b. If the sum of the squares of a and b is 34, then find the product of a and b.
Solution:
Given, a is 2 more than b
⇒ a = b + 2
And, sum of squares of a and b is 34
⇒ a2 + b2 = 34
Let’s replace a = (b + 2) in the above equation and solve for b
Then,
(b + 2)2Â + b2Â = 34
2b2 + 4b – 30 = 0
b2 + 2b – 15 = 0
(b + 5) (b – 3) = 0
So,
b = -5 or 3
Now,
For b = -5, a =-5 + 2 = -3
For b = 3, a = 3 + 2 = 5
Thus, the product of a and b is 15 in both cases.
Â
16. The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.
Solution:
Let’s assume the two positive numbers as a and b
Given, the difference between them is 5 and the sum of their squares is 73
So, we have
a – b = 5 … (i) and
a2 + b2 = 73 … (ii)
On squaring (i) on both sides, we get
(a – b)2 = 52
(a2 + b2) – 2ab = 25
73 – 2ab = 25 … [Using (ii), given]
So,
2ab = 73 – 25 = 48
ab = 24Â
Therefore, the product of numbers is 24.
Exercise 4(B)
1. Find the cube of:
(i) 3a – 2b
(ii) 5a + 3b
(iii)Â 2a + 1/2a
(iv) 3a – 1/a (a ≠0)Â
Solution:
Using the identities,
(a + b)3 = a3 + 3ab (a + b) + b3 and
(a – b)3 = a3 – 3ab (a – b) + b3
(i) (3a – 2b)3 = (3a)3 – 3 × 3a × 2b (3a – 2b) – (2b)3
= 27a3 – 18ab (3a – 2b) – 8b3
= 27a3 – 54a2b + 36ab2 – 8b3
(ii) (5a + 3b)3 = (5a)3 + 3 × 5a × 3b (5a + 3b) + (3b)3
= 125a3 + 45ab (5a + 3b) + 27b3
= 125a3 + 225a2b + 135ab2 + 27b3
(iii) (2a + 1/2a)3 = (2a)3 + 3 × 2a × 1/2a (2a + 1/2a) + (1/2a)3
= 8a3 + 3 (2a + 1/2a) + 1/8a3
= 8a3 + 6a + 3/2a + 1/8a3
(iv) (3a – 1/a)3 = (3a)3 – 3 × 3a × 1/a (3a – 1/a) – (1/a)3
= 27a3 – 9 (3a – 1/a) – 1/a3
= 27a3 – 27a + 9a – 1/a3
2. If a2 + 1/a2 = 47 and a ≠0 find:
(i)Â a + 1/a
(ii)Â a3 + 1/a3
Solution:
(i) Given, a2Â +Â 1/a2 = 47
We know that,
(a + 1/a)2 = a2Â +Â 1/a2 + 2 x a x 1/a
= (a2Â +Â 1/a2) + 2
= 47 + 2
= 49
So,
a + 1/a = √49
= ±7 … (1)
(ii) Using the identity
(a + b)3 = a3 + 3ab (a + b) + b3
Now,
(a + 1/a)3 = a3 + 1/a3 + 3(a + 1/a)
a3 + 1/a3 = (a + 1/a)3 – 3(a + 1/a)
= (±7)3 – 3(±7) … [From (1)]
= ±343 – ±21
Hence, a3 + 1/a3 = ±322
3. If a2 + 1/a2 = 18; a ≠0 find:
(i) a – 1/aÂ
(ii) a3 – 1/a3Â
Solution:
(i) Given, a2Â +Â 1/a2 = 18
Using the identity (a + b)2 = a2 + b2 + 2ab
Now,
(a – 1/a)2 = a2 + 1/a2 – 2(a)(1/a)
= (a2 + 1/a2) – 2
= 18 – 2
= 16
Hence,
a – 1/a = √16
= ±4 … (1)
(ii) Using the identity,
(a – b)3 = a3 – 3ab (a – b) + b3
Now,
(a – 1/a)3 = a3 – 3a(1/a) (a – 1/a) + (1/a)3
= a3 – 3 (a – 1/a) + 1/a3
a3 + 1/a3 = (a – 1/a)3 + 3 (a – 1/a)
= (±4)3 + 3(±4)
= ±64 ± 12
Hence,
a3 + 1/a3 = ±76
4. If a + 1/a = p and a ≠0; then show that:
a3 + 1/a3 = p (p2 – 3)
Solution:
Given, a + 1/a = p … (1)
Now, cubing on both sides
(a + 1/a)3 = p3
a3 + 1/a3 + 3(a + 1/a) = p3
a3 + 1/a3 = p3 – 3(a + 1/a)
= p3 – 3(p) [From (1)]
= p (p2 – 3)
– Hence proved
5. If a + 2b = 5; then show that:
a3Â + 8b3Â + 30ab = 125.
Solution:
Given, a + 2b = 5
Let’s cube it on both sides,
(a + 2b)3 = 53
a3 + 3(a)(2b)(a + 2b) + (2b)3 = 125
a3 + 6ab(a + 2b) + 8b3 = 125
a3 + 8b3 = 125 – 6ab (a + 2b)
= 125 – 6ab (5) … [Given]
= 125 – 30ab
So,
a3 + 8b3 + 30ab = 125
– Hence showed
6. If (a + 1/a)2 = 3 and a ≠0, then show: a3 + 1/a3 = 0.
Solution:
Given, (a + 1/a)2 = 3
⇒ a + 1/a = ±√3 … (1)
We know the identity,
(a + 1/a)3 = a3 + 1/a3 + 3(a + 1/a)
a3 + 1/a3 = (a + 1/a)3 – 3(a + 1/a)
= (±√3)3 – 3(±√3)
= ±3√3 – (±3√3)
= 0
Thus, a3 + 1/a3 = 0
Â
7. If a + 2b + c = 0; then show that:
a3Â + 8b3Â + c3Â = 6abc
Solution:
We have, a + 2b + c = 0
a + 2b = -c
Now, on cubing it on both sides we get
(a + 2b)3 = (-c)3
a3 + (2b)3 + 3(a)(2b)(a + 2b) = -c3
a3 + 8b3 + 6ab (a + 2b) = -c3
a3 + 8b3 + 6ab (-c) = -c3
a3 + 8b3 – 6abc = -c3
Hence,
a3 + 8b3 + c3 = 6abc
8. Use property to evaluate:
(i) 133Â + (-8)3Â + (-5)3
(ii) 73Â + 33Â + (-10)3Â
(iii) 93 – 53 – 43
(iv) 383Â + (-26)3Â + (-12)3
Solution:
The property is if a + b + c = 0 then
a3Â + b3Â + c3Â = 3abc
Now,
(i) a = 13, b = -8 and c = -5
⇒ 133 + (-8)3 + (-5)3 = 3(13) (-8) (-5) … [Since, 13 + (-8) + (-5) = 0]
= 1560
Â
(ii) a = 7, b = 3, c = -10
⇒ 73 + 33 + (-10)3 = 3(7) (3) (-10) … [Since, 7 + 3 + (-10) = 0]
= -630
(iii)a = 9, b = -5, c = -4
⇒ 93 – 53 – 43 = 93 + (-5)3 + (-4)3 … [Since, 9 + (-5) + (-4) = 0]
= 3(9) (-5) (-4) = 540
Â
(iv) a = 38, b = -26, c = -12
⇒ 383 + (-26)3 + (-12)3 = 3(38) (-26) (-12) … [Since, 38 + (-26) + (-12) = 0]
= 35568
Â
9. If a ≠0 and a – 1/a = 3; find:
(i) a2 + 1/a2
(ii) a3 – 1/a3
Solution:
(i) We have, a – 1/a = 3
On squaring on both sides, we get
(a – 1/a)2 = 32
a2 + 1/a2 – 2 = 9
a2 + 1/a2 = 9 + 2
Hence,
a2 + 1/a2 = 11
Â
(ii) We have, a – 1/a = 3
On cubing on both sides, we get
(a – 1/a)3 = 33
a2 – 1/a3 – 3(a – 1/a) = 27
a2 – 1/a3 = 27 + 3(a – 1/a)
= 27 + 3(3)
= 27 + 9
Hence,
a3 – 1/a3 = 36
10. If a ≠0 and a – 1/a = 4; find:
(i)Â a2 + 1/a2
(ii)Â a4 + 1/a4
(iii) a3 – 1/a3
Solution:
(i) We have, a – 1/a = 4 … (a)
On squaring it on both sides, we get
(a – 1/a)2 = 42
a2 + 1/a2 – 2(a)(1/a) = 16
a2 + 1/a2 – 2 = 16
a2 + 1/a2 = 16 + 2 = 18 … (1)
Hence, a2 + 1/a2 = 18
(ii) Now, we know that
a4 + 1/a4 = (a2 + 1/a2)2 – 2
= 182 – 2 … [From (1)]
= 324 – 2
Hence, a4 + 1/a4 = 322
(iii) On cubing (i) on both sides, we get
(a – 1/a)3 = 43
a3 – 1/a3 – 3(a – 1/a) = 64
a3 – 1/a3 = 64 + 3(a – 1/a)
= 64 + 3(4) … [Given]
= 64 + 12
Hence, a3 – 1/a3 = 76
11. If x ≠0 and x + 1/x = 2; then show that:
x2 + 1/x2 = x3 + 1/x3 = x4 + 1/x4
Solution:
We have, x +Â 1/x = 2
We know that,
(x +Â 1/x)2 = x2 +Â 1/x2 + 2
(2)2 = x2 +Â 1/x2 + 2
x2 + 1/x2 = 4 – 2
= 2 … (i)
Next, calculating
(x + 1/x)3 = x3 + 1/x3 + 3(x + 1/x)
(2)3 = x3 + 1/x3 + 3(2)
x3 + 1/x3 = 23 – 3(2)
= 8 – 6
= 2 … (ii)
Next, we know that
x4 + 1/x4 = (x2 + 1/x2) – 2
= 22 – 2 … [From (i)]
= 4 – 2
= 2 … (iii)
Therefore, from (i), (ii) and (iii) we have
x2 + 1/x2 = x3 + 1/x3 = x4 + 1/x4
Â
12. If 2x – 3y = 10 and xy = 16; find the value of 8x3 – 27y3.
Solution:
Given,Â
2x – 3y = 10 … (i) andÂ
xy = 16 … (ii)
Now, on cubing (i) on both sides
(2x – 3y)3 = 103
(2x)3 – 3(2x)(3y) (2x – 3y) – (3y)3 = 1000 []
8x3 – 18(xy) (2x – 3y) – 27y3 = 1000
8x3 – 18 × 16 × 10 – 27y3 = 1000
8x3 – 2880 – 27y3 = 1000
8x3 – 27y3 = 1000 + 2880
8x3 – 27y3 = 3880
13. Expand:
(i) (3x + 5y + 2z) (3x – 5y + 2z)
(ii) (3x – 5y – 2z) (3x – 5y + 2z)
Solution:
(i) We have, (3x + 5y + 2z) (3x – 5y + 2z)
= {(3x + 2z) + (5y)} {(3x + 2z) – (5y)} … [By grouping]
= (3x + 2z)2 – (5y)2 … [As (a + b) (a – b) = a2 – b2]
= 9x2 + 4z2 + (2 × 3x × 2z) – 25y2
= 9x2 + 4z2 + 12xz – 25y2
= 9x2 + 4z2 – 25y2 + 12xz
Â
(ii) We have, (3x – 5y – 2z) (3x – 5y + 2z)
= {(3x – 5y) – (2z)} {(3x – 5y) + (2z)} … [By grouping]
= (3x – 5y)2 – (2z)2 … [As (a + b) (a – b) = a2 – b2]
= 9x2 + 25y2 – 2 × 3x × 5y – 4z2
= 9x2 + 25y2– 30xy – 4z2
= 9x2 +25y2 – 4z2 – 30xy
Â
14. The sum of two numbers is 9 and their product is 20. Find the sum of their
(i) Squares (ii) Cubes
Solution:
Given, the sum of two numbers is 9 and their product is 20
Let’s assume the numbers to ‘a’ and ‘b’
So, we have
a + b = 9 … (1) and
ab = 20 … (2)
Now,
On squaring (1) on both sides gives, we get
(a + b)2Â = 92
a2Â + b2Â + 2ab = 81
a2 + b2 + 2(20) = 81 … [From (2)]
a2Â + b2Â + 40 = 81
a2 + b2 = 81 – 40 = 41
(i) Hence, the sum of their squares is 41
Next,
On cubing (1) on both sides, we get
(a + b)3Â = 93
a3Â + b3Â + 3ab (a + b) = 729
a3 + b3 + 3 × (20) × (9) = 729 … [From (1) and (2)]
a3 + b3 = 729 – 540 = 189
(ii) Hence, the sum of their cubes is 189.
Â
15. Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:
(i) Sum of these numbers
(ii) Difference of their cubes
(iii) Sum of their cubes.
Solution:
Given x – y = 5 and xy = 24 (x>y)
(x + y)2 = (x – y)2 + 4xy = 25 + 96 = 121
So, x + y = 11; sum of these numbers is 11.
Cubing on both sides gives
(x – y)3 = 53
x3 – y3 – 3xy(x – y) = 125
x3 – y3 – 72(5) = 125
x3 – y3= 125 + 360 = 485
So, difference of their cubes is 485.
Cubing both sides, we get
(x + y)3Â = 113
x3Â + y3Â + 3xy(x + y) = 1331
x3 + y3 = 1331 – 72(11) = 1331 – 792 = 539
So, sum of their cubes is 539.
Â
16. If 4x2 + y2 = a and xy = b, find the value of 2x + y.
Solution:
Given, xy = b … (i) and 4x2 + y2 = a … (ii)
Now,
(2x + y)2Â = (2x)2Â + 4xy + y2
= (4x2Â + y2)Â + 4xy
= a + 4b … [Using (i) and (ii)]
Hence,
2x + y = ±√(a + 4b)
Exercise 4(C)
1. Expand:
(i) (x + 8) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (x – 8) (x + 10)
(iv) (x – 8) (x – 10)Â
Solution:
Using the identity, (x + a) (x + b) = x2 + (a + b) x + ab
(i) We have, (x + 8) (x + 10)
= x2 + (8 + 10) x + 8 × 10
= x2 + 18x + 80
(ii) We have, (x + 8) (x – 10)
= x2 + (8 – 10) x + 8 × (-10)
= x2 – 2x – 80
(iii) (ii) We have, (x – 8) (x + 10)
= x2 + (-8 + 10) x + (-8) × 10
= x2 + 2x – 80
(iv) We have, (x – 8) (x – 10)
= x2 + (-8 – 10) x + (-8) × (-10)
= x2 – 18x + 80
2. Expand:
(i) (2x – 1/x) (3x + 2/x)
(ii) (3a + 2/b) (2a – 3/b)
Solution:
(i) We have, (2x – 1/x) (3x + 2/x)
= (2x)(3x) + (2x)(2/x) – (1/x)(3x) – (1/x)(2/x)
= 6x2 + 4 – 3 – 2/x2
= 6x2 + 1 – 2/x2
(ii) We have, (3a + 2/b) (2a – 3/b)
= (3a)(2a) – (3a)(3/b) + (2/b)(2a) – (2/b)(3/b)
= 6a2 – 9a/b + 4a/b – 6/b2
= 6a2 – 5a/b – 6/b2
3. Expand:
(i) (x + y – z)2
(ii) (x – 2y + 2)2
(iii) (5a – 3b + c)2
(iv) (5x – 3y – 2)2
(v) (x – 1/x + 5)2
Solution:
(i) (x + y – z)2 = x2 + y2 + z2 + 2(x)(y) – 2(y)(z) – 2(z)(x)
= x2 + y2 + z2 + 2xy – 2yz – 2zx
(ii) (x – 2y + 2)2 = x2 + (-2y)2 + 22 + 2(x)(-2y) + 2(-2y)(2) + 2(2)(x)
= x2 + 4y2 + 4 – 4xy – 8y + 4x
(iii) (5a – 3b + c)2 = (5a)2 + (-3b)2 + c2 + 2(5a)(-3b) + 2(-3b)(c) + 2(c)(5a)
= 25a2 + 9b2 + c2 – 30ab – 6bc + 10ac
(iv) (5x – 3y – 2)2 = (5x)2 + (-3y)2 + (-2)2 + 2(5x)(-3y) + 2(-3y)(-2) + 2(-2)(5x)
= 25x2 + 9y2 + 4 – 30xy + 12y – 20x
(v) (x – 1/x + 5)2 = (x)2 + (-1/x)2 + (5)2 + 2(x)(-1/x) + 2(-1/x)(5) + 2(5)(x)
= x2 + 1/x2 + 25 – 2 – 10/x + 10x
= x2 + 1/x2 + 23 – 10/x + 10x
Â
4. If a + b + c = 12 and a2Â + b2Â + c2Â = 50; find ab + bc + ca.
Solution:
Given, a + b + c = 12 and a2Â + b2Â + c2Â = 50
We know that,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
122 = 50 + 2(ab + bc + ca)
144 = 50 + 2(ab + bc + ca)
ab + bc + ca = (144 – 50)/ 2
= 94/2
Thus,
ab + bc + ca = 47
5. If a2 + b2 + c2 = 35 and ab + bc + ca = 23; find a + b + c.
Solution:
Given, a2Â + b2Â + c2Â = 35 and ab + bc + ca = 23
We know that,
(a + b + c)2 = (a2 + b2 + c2) + 2(ab + bc + ca)
(a + b + c)2 = 35 + 2(23)
(a + b + c)2 = 35 + 46
(a + b + c)2 = 81
(a + b + c) = ±√81
Thus,
a + b + c = ±9
6. If a + b + c = p and ab + bc + ca = q; find a2Â + b2Â + c2.
Solution:
Given, a + b + c = p and ab + bc + ca = q
We know that,
(a + b + c)2 = (a2 + b2 + c2) + 2(ab + bc + ca)
(p)2 = (a2 + b2 + c2) + 2(q)
⇒ a2 + b2 + c2 = p2 – 2q
7. If a2 + b2 + c2 = 50 and ab + bc + ca = 47, find a + b + c.
Solution:
Given, a2 + b2 + c2 = 50 and ab + bc + ca = 47
We know that,
(a + b + c)2 = (a2 + b2 + c2) + 2(ab + bc + ca)
(a + b + c)2 = 50 + 2(47)
(a + b + c)2 = 50 + 94
= 144
⇒ (a + b + c) = √144
Thus,
a + b + c = ±12
8. If x + y – z = 4 and x2 + y2 + z2 = 30, then find the value of xy – yz – zx.
Solution:
Given, x + y – z = 4 and x2 + y2 + z2 = 30
We know that,
(x + y – z)2 = x2 + y2 + z2 + 2(xy – yz – zx)
42 = 30 + 2(xy – yz – zx)
16 – 30 = 2(ab + bc + ca)
xy – yz – zx = -14/ 2
Thus,
xy – yz – zx = -7
Exercise 4(D)
1. If x + 2y + 3z = 0 and x3Â + 4y3Â + 9z3Â = 18xyz; evaluate:
Solution:
Given, x3 + 4y3 + 9z3 = 18xyz and x + 2y + 3z = 0
So,
x + 2y = – 3z, 2y + 3z = -x and 3z + x = -2y
Now,
2. If a + 1/a = m and a ≠0; find in terms of ‘m’; the value of:
(i) a – 1/a
(ii) a2 – 1/a2
Solution:
(i) Given, a +Â 1/a = m
On squaring on both sides, we get
(a +Â 1/a)2 = m2
a2 + 1/a2 + 2 = m2
a2 + 1/a2 = m2 – 2 … (1)
Now, consider the expansion
(a – 1/a)2 = a2 + 1/a2 – 2
= m2 – 2 – 2 … [From (1)]
= m2 – 4
So,
(a – 1/a) = ±√(m2 – 4) … (2)
(ii) We know that,
a2 – 1/a2 = (a – 1/a) (a + 1/a)
= m [±√(m2 – 4)]
= ±m√(m2 – 4)
Â
3. In the expansion of (2x2 – 8) (x – 4)2; find the value of
(i) coefficient of x3
(ii) coefficient of x2
(iii) constant term
Solution:
We have, (2x2 – 8) (x – 4)2
= (2x2 – 8) (x2 – 2 × 4 × x + 42)
= (2x2 – 8) (x2 – 8x + 16)
= 2x2(x2 – 8x + 16) – 8(x2 – 8x + 16)
= 4x4 – 16x3 + 32x2 – 8x2 + 64x – 128
= 4x4 – 16x3 + 24x2 + 64x – 128
Now,
(i) coefficient of x3 = -16
(ii) coefficient of x2Â = 24
(iii) constant term = -128
4. If x > 0 and x2 + 1/9x2 = 25/36. Find: x3 + 1/27x3
Solution:
Given, x2 + 1/9x2 = 25/36 … (1)
Now, consider the expansion
(x + 1/3x)2 = x2 + (1/3x)2 + (2 × x × 1/3x)
= (x2 + 1/9x2) + 2/3
= 25/36 + 2/3 … [From (1)]
= 49/36
So,
(x + 1/3x) = ±√(49/36)
= ±7/6 … (2)
Now, consider the expansion
(x + 1/3x)3 = x3 + (1/3x)3 + 3(x + 1/3x)
(7/6)3 = x3 + (1/3x)3 + 3(7/6) …… [From (2)]
343/216 = x3 + 1/27x3 + 21/6
x3 + 1/27x3 = 343/216 – 21/6
= (343 – 252)/216
= 91/216
Thus, x3 + 1/27x3 = 91/216
5. If 2(x2Â + 1) = 5x, find:
(i) x – 1/xÂ
(ii) x3 – 1/x3
Solution:
(i) Given, 2(x2Â + 1) = 5x
x2Â + 1 = 5x/2
On dividing by x on both sides, we have
(x2 + 1)/x = 5/2
⇒ (x + 1/x) = 5/2 … (1)
Â
Now, consider the expansion of (x + 1/x)2
(x + 1/x)2 = x2 + 1/x2 + 2
(5/2)2 = x2 + 1/x2 + 2 … [From (1)]
x2 + 1/x2 = 25/4 – 2
= (25 – 8)/4
= 17/4 … (2)
Now,
(x – 1/x)2 = x2 + 1/x2 – 2
= 17/4 – 2 … [From (2)]
= (17 – 8)/4
= 9/4
So,
x – 1/x = √9/4
Thus,
(i) x – 1/x = ±3/2 … (3)
Â
Next, we know that
(x3 – 1/x3) = (x – 1/x)3 + 3(x – 1/x)
= (±3/2)3 + 3(±3/2) … [From (3)]
= ± 27/8 ± 9/2
= ± (27 + 36)/8
= ± 63/8
(ii) Thus, x3 – 1/x3 = ±63/8
Â
6. If a2 + b2 = 34 and ab = 12; find:
(i) 3(a + b)2 + 5(a – b)2
(ii) 7(a – b)2 – 2(a + b)2
Solution:
We have, a2Â +Â b2Â = 34 and ab= 12
We know that,
(a + b)2 = (a2 + b2) + 2ab
= 34Â +Â 2 x 12
= 34Â +Â 24
= 58Â Â
Also, we know that
(a – b)2 = (a2 + b2) – 2abÂ
= 34 – 2 x 12
= 34- 24
= 10
(i) 3(a + b)2 + 5(a – b)2
= 3 x 58Â +Â 5 x 10
= 174Â +Â 50
= 224
Â
(ii) 7(a – b)2 – 2(a + b)2
= 7 x 10 – 2 x 58
= 70 – 116
= -46
Â
7. If 3x – 4/x = 4 and x ≠0; find: 27x3 – 64/x3.
Solution:
Given, 3x – 4/x = 4
Now, let’s consider the expansion of (3x – 4/x)3
(3x – 4/x)3 = 27x3 – 64/x3 – 3 × 3x × 4/x(3x – 4/x)
(4)3 = 27x3 – 64/x3 – 36(3x – 4/x)
64 = 27x3 – 64/x3 – 36(4)
64 = 27x3 – 64/x3 – 144
27x3 – 64/x3 = 144 + 64
Hence,
27x3 – 64/x3 = 208
8. If x2 + 1/x2 = 7 and x ≠0; find the value of: 7x3 + 8x – 7/x3 – 8/x.
Solution:
Given, x2 + 1/x2 = 7
On subtracting 2 from both sides, we get
x2 + 1/x2 – 2 = 7 – 2
(x – 1/x)2 = 5
x – 1/x = ±√5 … (1)
Now, consider
(x – 1/x)3 = x3 – 1/x3 – 3(x – 1/x)
(±√5)3 = x3 – 1/x3 – 3(±√5)
x3 – 1/x3 = (±√5)3 + 3(±√5) … (2)
Taking,
7x3 + 8x – 7/x3 – 8/x
= 7x3 – 7/x3 + 8x – 8/x
= 7(x3 – 1/x3) + 8(x – 1/x)
= 7[(±√5)3 + 3(±√5)] + 8(±√5)
= ±35√5 ± 21√5 ± 8√5
= ±64√5
9. If x = 1/(x – 5) and x ≠5, find x2 – 1/x2.
Solution:
Given, x = 1/(x – 5)
By cross multiplying, we have
x (x – 5) = 1Â
x2 – 5x = 1
x2 – 1 = 5x
Dividing both sides by x,
(x2 – 1)/x = 5
(x – 1/x) = 5 … (1)
Now,
(x – 1/x)2 = 52
x2 + 1/x2 – 2 = 25
x2 + 1/x2 = 25 + 2
= 27 … (2)
Considering the expansion (x + 1/x)2
(x + 1/x)2 = x2 + 1/x2 + 2
(x + 1/x)2 = 27 + 2 … [From (1)]
(x + 1/x)2 = 29
x + 1/x = ±√29 … (3)
We know that,
x2 – 1/x2 = (x + 1/x) (x – 1/x)
= (±√29) (5) … [From (3)]
= ±5√29
10. If x = 1/(5 – x) and x ≠5; find x3 + 1/x3.
Solution:
Given, x = 1/(5 – x)
By cross multiplying, we have
x (5 – x) = 1
x2 – 5x = -1
x2Â +Â 1Â =Â 5x
Dividing both sides by x,
(x2 + 1)/x = 5
x + 1/x = 5 … (1)
Now,
(x + 1/x)3 = x3 + 1/x3 + 3(x + 1/x)
x3 + 1/x3 = (x + 1/x)3 – 3(x + 1/x)
= 53 – 3(5)
= 125 – 15
= 110
Thus, x3 + 1/x3 = 110
11. If 3a + 5b + 4c = 0,
Show that: 27a3Â + 125b3Â + 64c3Â = 180abc
Solution:
Given, 3a + 5b + 4c = 0
⇒ 3a + 5b = -4c
On cubing on both sides, we have
(3a + 5b)3 = (-4c)3
(3a)3 + (5b)3 + 3 x 3a x 5b (3a + 5b) = -64c3
27a3 + 125b3 + 45ab (-4c) = -64c3
27a3 + 125b3 – 180abc = -64c3
27a3Â +Â 125b3Â +Â 64c3Â =Â 180abcÂ
– Hence Proved.
12. The sum of two numbers is 7 and the sum of their cubes is 133, find the sum of their square.
Solution:
Let’s assume a and b to be the two numbers
So, a + b = 7 and a3 + b3 = 133
We know that,
(a + b)3 = a3 + b3 + 3ab (a + b)
(7)3Â = 133Â +Â 3ab (7)
343Â =Â 133Â +Â 21ab
21ab = 343 – 133
= 210
⇒ ab = 21
Now,Â
a2 + b2 = (a + b)2 – 2ab
= 72 – 2 x 10Â
= 49 – 20Â
=Â 29
13. In each of the following, find the value of ‘a’:
(i) 4x2Â + ax + 9 = (2x + 3)2
(ii) 4x2 + ax + 9 = (2x – 3)2
(iii) 9x2 + (7a – 5)x + 25 = (3x + 5)2
Solution:
(i) 4x2Â + ax + 9 = (2x + 3)2 = 4x2 + 12x + 9
On comparing coefficients of x terms, we get
ax = 12x
So,
a = 12
(ii) 4x2 + ax + 9 = (2x – 3)2 = 4x2 + 12x + 9
On comparing coefficients of x terms, we get
ax = -12x
So,
a = -12
(iii) 9x2 + (7a – 5)x + 25 = (3x + 5)2 = 9x2 + 30x + 25
On comparing coefficients of x terms, we get
(7a – 5)x = 30x
7a – 5 = 30
7a = 35
⇒ a = 5
Â
14. If (x2 + 1)/x = 3 1/3 and x > 1; find
(i) x – 1/xÂ
(ii) x3 – 1/x3
Solution:
Given,
(x2 + 1)/x = 3 1/3 = 10/3
x + 1/x = 10/3
On squaring on both sides, we get
(x + 1/x)2 = (10/3)2
x2 + 1/x2 + 2 = 100/9
x2 + 1/x2 = 100/9 – 2
= (100 – 18)/9
= 82/9
Now,
(x – 1/x)2 = x2 + 1/x2 – 2
= 82/9 – 2
= (82 – 18)/9
= 64/9
x – 1/x = √(64/9)
= ±8/3
On cubing both sides, we get
(x – 1/x)3 = (8/3)3
x3 – 1/x3 – 3(x – 1/x) = 512/27
x3 – 1/x3 = 3(x – 1/x) + 512/27
= 3(8/3) + 512/27
= 24/3 + 512/27
= (216 + 512)/27
= 728/27
Therefore, x3 – 1/x3 = 728/27
15. The difference between two positive numbers is 4 and the difference between their cubes is 316.
Find:
(i) Their product
(ii) The sum of their squares
Solution:
Given, difference between two positive numbers is 4
And, the difference between their cubes is 316
Let’s assume the positive numbers to be a and b
So,
a – b = 4
a3 – b3 = 316
On cubing both sides, we have
(a – b)3 = 64
a3 – b3 – 3ab(a – b) = 64
Also,
Given: a3 – b3 = 316
So,
316 – 64 = 3ab(4)
252 = 12ab
So,
ab = 21
Thus, the product of numbers is 21
Now,
On squaring both sides, we get
(a – b)2 = 16
a2 + b2 – 2ab = 16
a2Â + b2Â = 16 + 42 = 58
Thus, sum of their squares is 58.
Exercise 4(E)
1. Simplify:
(i) (x + 6)(x + 4)(x – 2)
(ii) (x – 6)(x – 4)(x + 2)
(iii) (x – 6)(x – 4)(x – 2)
(iv) (x + 6)(x – 4)(x – 2)Â
Solution:
Using identity:
(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
(i) We have, (x + 6)(x + 4)(x – 2)
= x3 + (6 + 4 – 2)x2 + [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2)
= x3 + 8x2 + (24 – 8 – 12)x – 48
= x3 + 8x2 + 4x – 48
(ii) We have, (x – 6)(x – 4)(x + 2)
= x3 + (-6 – 4 + 2)x2 + [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2
= x3 – 8x2 + (24 – 8 – 12)x + 48
= x3 – 8x2 + 4x + 48
(iii) We have, (x – 6)(x – 4)(x – 2)
= x3 + (-6 – 4 – 2)x2 + [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2)
= x3 – 12x2 + (24 + 8 + 12)x – 48
= x3 – 12x2 + 44x – 48
Â
(iv) We have, (x + 6)(x – 4)(x – 2)
= x3 + (6 – 4 – 2)x2 + [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2)
= x3 – 0x2 + (-24 + 8 – 12)x + 48
= x3 – 28x + 48Â
2. Simply using following identity:
(a ± b) (a2 ∓ ab + b2) = a3 ± b3
(i) (2x + 3y) (4x2 – 6xy + 9y2)
(ii) (3x – 5/x) (9x2 + 15 + 25/x2)
(iii) (a/3 – 3b) (a2 + ab + 9b2)
Solution:
(i) We have, (2x + 3y) (4x2 – 6xy + 9y2)
= (2x + 3y) [(2x)2 – (2x)(3y) + (3y)2]
= (2x)3 + (3y)3
= 8x3 + 27y3
(ii) We have, (3x – 5/x) (9x2 + 15 + 25/x2)
= (3x – 5/x) [(3x)2 + (3x)(5/x) + (5/x)2]
= (3x)3 + (5/x)3
= 27x3 + 125/x3
(iii) We have, (a/3 – 3b) (a2/9 + ab + 9b2)
= (a/3 – 3b) [(a/3)2 + (a/3)(3b) + (3b)2]
= (a/3)3 – (3b)3
= a3/27 – 27b3
Â
3. Using suitable identity, evaluate
(i) (104)3
(ii) (97)3Â
Solution:
Using identity: (a ± b)3 = a3 ± b3 ± 3ab(a ± b)
(i) (104)3Â = (100 + 4)3
= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)
= 1000000 + 64 + 1200 × 104
= 1000000 + 64 + 124800
= 1124864
Â
(ii) (97)3 = (100 – 3)3
= (100)3 – (3)3 – 3 × 100 × 3(100 – 3)
= 1000000 – 27 – 900 × 97
= 1000000 – 27 – 87300
= 912673
4. Simply:
Solution:
We know that,
If a + b + c = 0, then a3 + b3 + c3 = 3abc
Now, if
(x2 – y2) + (y2 – z2) + (z2 – x2) = 0
Then, we have
(x2 – y2)3 + (y2 – z2)3 + (z2 – x2)3 = 3(x2 – y2)(y2 – z2)(z2 – x2) … (1)
Similarly, if
x – y + y – z + z – x = 0
Then,
(x – y)3 + (y – z)3 + (z – x)3 = 3(x – y)(y – z)(z – x) … (2)
Now,
= (x + y)(y + z)(z + x)
Â
5. Evaluate:
Solution:
(i) We have,
= a + b
= 0.8 + 0.5
= 1.3
(ii) We have,
Â
6. If a – 2b + 3c = 0; state the value of a3 – 8b3 + 27c3.
Solution:
Given, a – 2b + 3c = 0
Then,
a3 – 8b3 + 27c3 = a3 + (-2b)3 + (3c)3 = 3(a)( -2b)(3c)
= -18abcÂ
7. If x + 5y = 10; find the value of x3 + 125y3 + 150xy – 1000.
Solution:
Given, x + 5y = 10
On cubing both sides, we get
(x + 5y)3Â = 103
x3Â + (5y)3Â + 3(x)(5y)(x + 5y) = 1000
x3Â + (5y)3Â + 3(x)(5y)(10) = 1000
x3Â + (5y)3Â + 150xy = 1000
Thus,
x3 + (5y)3 + 150xy – 1000 = 0Â
8. If x = 3 + 2√2, find:
(i) 1/x
(ii) x – 1/x
(iii) (x – 1/x)3
(iv) x3 – 1/x3
Solution:
We have, x = 3 + 2√2
(i) 1/x = 1/(3 + 2√2)
= (3 – 2√2)/ [(3 + 2√2) × (3 – 2√2)]
= (3 – 2√2)/ [32 – (2√2)2]
= (3 – 2√2)/ (9 – 8)
= 3 – 2√2
(ii) x – 1/x = (3 + 2√2) – (3 – 2√2) … [From (i)]
= (3 + 2√2 – 3 + 2√2)
= 4√2
(iii) (x – 1/x)3 = (4√2)3 … [From (ii)]
= (64 x 2√2)
= 128√2
(iv) (x3 – 1/x3) = (x – 1/x)3 – 3(x – 1/x)
= 128√2 – 3(4√2) … [From (iii) and (ii)]
= 128√2 – 12√2Â
9. If a + b = 11 and a2Â + b2Â = 65; find a3Â + b3.
Solution:
Given, a + b = 11 and a2Â + b2Â = 65
Now, we know that
(a + b)2 = a2 + b2 + 2ab
(11)2 = 65 + 2ab
121 = 65 + 2ab
2ab = 121 – 65
ab = (121 – 65)/2
= 56/2
= 28
Considering the expansion (a3 + b3)
(a3 + b3) = (a + b) (a2 + b2 – ab)
= (11) (65 – 28)
= 11 × 37
= 407
Thus, a3Â + b3 = 407
10. Prove that:
x2+ y2 + z2 – xy – yz – zx is always positive.
Solution:
We have, x2 + y2 + z2 – xy – yz – zx
= 2(x2 + y2 + z2 – xy – yz – zx)
= 2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx
= x2 + x2 + y2 + y2 + z2 + z2 – 2xy – 2yz – 2zx
= (x2 + y2 – 2xy) + (z2 + x2 – 2zx) + (y2 + z2 – 2yz)
= (x – y)2 + (z – x)2 + (y – z)2
As the square of any number is positive, the given equation is always positive.
11. Find:
(i)Â (a + b)(a + b)
(ii)Â (a + b)(a + b)(a + b)
(iii) (a – b)(a – b)(a – b) by using the result of part (ii)
Solution:
(i)Â We have, (a + b)(a + b)
= (a + b)2
= a × a + a × b + b × a + b × b
= a2 + ab + ab + b2
= a2Â + b2Â + 2ab
Â
(ii)Â We have, (a + b)(a + b)(a + b)
= (a × a + a × b + b × a + b × b)(a + b)
= (a2 + ab + ab + b2)(a + b)
= (a2Â + b2Â + 2ab)(a + b)
= a2 × a + a2 × b + b2 × a + b2 × b + 2ab × a + 2ab × b
= a3Â + a2Â b + ab2Â + b3Â + 2a2b + 2ab2
= a3Â + b3Â + 3a2b + 3ab2
Â
(iii) We have, (a – b)(a – b)(a – b)
In result (ii), replacing b by -b, we get (a – b)(a – b)(a – b)
= a3Â + (-b)3Â + 3a2(-b) + 3a(-b)2
= a3 – b3 – 3a2b + 3ab2
Selina Solutions for Class 9 Maths Chapter 4- Expansions
The Chapter 4, Expansions, contains 5 exercises and the Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.
4.1 Introduction
4.2 Identities
4.3 Expansions of (a+b)3
4.4 Expansions of (x+a)(x+b)
4.5 Expansions of (a+b+c)2
4.6 Using Expansions
4.7 Special Products
Selina Solutions for Class 9 Maths Chapter 4- Expansions
In chapter 4 of Class 9, the students are taught about a process in which the contents of brackets are evaluated, the expansion. The chapter helps the students in recollecting the identities that were taught in earlier classes. An identity is an equation that is true for all values of its variables. Study the Chapter 4 of Selina textbook to understand more about Expansions. Learn the Selina Solutions for Class 9 effectively to come out with flying colours in the examinations.
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