यदि $x+\frac{1}{x}=3$, तो $\left(x^{5}+\frac{1}{x^{5}}\right)$ का मूल्य है

$\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2$

$\Rightarrow x^{2}+\frac{1}{x^{2}}=9-2=7$

फिर से,

$\left(x+\frac{1}{x}\right)^{3}=x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right)$

$\Rightarrow 27=x^{3}+\frac{1}{x^{3}}+3 \times 3$

$\Rightarrow x^{3}+\frac{1}{x^{3}}=18$

$\therefore \quad\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{3}+\frac{1}{x^{3}}\right)$

$\quad=7 \times 18=126$

$\Rightarrow x^{5}+x+\frac{1}{x}+\frac{1}{x^{5}}=126$

$\Rightarrow x^{5}+\frac{1}{x^{5}}=126-3=123$