Question

If $x-\frac{1}{x}=8$, then find the value of $x^4+\frac{1}{x^4}$.

Answer 1

$x-\frac{1}{x}=8$,

Squaring, $x^2+\frac{1}{x^2}-2=64$

$\Rightarrow x^2+\frac{1}{x^2}=66$

Again squaring, $x^4+\frac{1}{x^4}+2=4356$

$\Rightarrow x^4+\frac{1}{x^4}=4354$.