If $x+\frac{1}{x}=3$ then find $x^5+\frac{1}{x^5}$.

Solve for $x^5+\frac{1}{x^5}$

It is given that $x+\frac{1}{x}=3$

Squaring on both sides

$$\begin{gathered} \Rightarrow {\left(x+\frac{1}{x}\right)}^2=9 \\ \Rightarrow x^2+\frac{1}{x^2}+2=9 \end{gathered}$$

$\Rightarrow x^2+\frac{1}{x^2}=7$…………..(1)

Similarly

$$\begin{gathered} \Rightarrow {\left(x+\frac{1}{x}\right)}^3=27 \\ \Rightarrow x^3+\frac{1}{x^3}+3x^2\frac{1}{x}+3x\frac{1}{x^2}=27 \\ \Rightarrow x^3+\frac{1}{x^3}+3x+3\frac{1}{x}=27 \\ \Rightarrow x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=27 \\ \Rightarrow x^3+\frac{1}{x^3}+3\times 3=27 \\ \Rightarrow x^3+\frac{1}{x^3}=27-9 \end{gathered}$$

$x^3+\frac{1}{x^3}=18$…………(2)

Multiply equation (1) and (2)

$$\begin{gathered} \Rightarrow \left(x^2+\frac{1}{x^2}\right)\times \left(x^3+\frac{1}{x^3}\right)=18\times 7 \\ \Rightarrow x^5+\frac{1}{x^5}+x+\frac{1}{x}=126 \\ \Rightarrow x^5+\frac{1}{x^5}=126-3 \\ \Rightarrow x^5+\frac{1}{x^5}=123 \end{gathered}$$

Hence the value of $x^5+\frac{1}{x^5}$is $123$