Question

If $x-\frac{1}{x}=8$, find the value of :
$x^2+\frac{1}{x^2}$

Answer 1

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

${(x-\frac{1}{x})}^2=64$

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

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

$\Rightarrow x^2+\frac{1}{x^2}=64+2=66$

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