If $x^2+\frac{1}{x^2}=51$, then the value of $x^3-\frac{1}{x^3}$ is :

The correct option is A $364$

Consider $x^2+\frac{1}{x^2}=51$ and solve it as shown below:

$x^2+\frac{1}{x^2}=51$

$\Rightarrow x^2+\frac{1}{x^2}=49+2$

$\Rightarrow x^2+\frac{1}{x^2}-2=49$

$\Rightarrow {(x-\frac{1}{x})}^2=49$ ...$[\because {(a-b)}^2=a^2+b^2-2ab]$

$\Rightarrow x-\frac{1}{x}=\sqrt{49}$

$\Rightarrow x-\frac{1}{x}=7.......\left(1\right)$

We know the formula $a^3-b^3=(a-b)(a^2+b^2+ab)$

Now substitute $a=x$ and $b=\frac{1}{x}$ in the above formula, then we get:

$a^3-b^3=(a-b)(a^2+b^2+ab)$

$\Rightarrow x^3-\frac{1}{x^3}=(x-\frac{1}{x})[x^2+\frac{1}{x^2}+(x\times \frac{1}{x})]$

$\Rightarrow x^3-\frac{1}{x^3}=(x-\frac{1}{x})[x^2+\frac{1}{x^2}+1]$

$\Rightarrow x^3-\frac{1}{x^3}=(x-\frac{1}{x})(51+1)$ ...$[\because x^2+\frac{1}{x^2}=51]$

$\Rightarrow x^3-\frac{1}{x^3}=52\times 7$ ...From eqn (1)

$\Rightarrow x^3-\frac{1}{x^3}=364$