Question

If $x-\frac{1}{x}=3$ ,then find the value of $x^4+\frac{1}{x^4}$

Answer 1

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

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

$\Rightarrow x^2+\frac{1}{x^2}={\left(3\right)}^2+2$

$\Rightarrow x^2+\frac{1}{x^2}=9+2=11$

$\Rightarrow {(x^2+\frac{1}{x^2})}^2=x^4+\frac{1}{x^4}+2x^2\times \frac{1}{x^2}$

$\Rightarrow {(x^2+\frac{1}{x^2})}^2=x^4+\frac{1}{x^4}+2$

$\Rightarrow x^4+\frac{1}{x^4}={(x^2+\frac{1}{x^2})}^2-2$

$\Rightarrow x^4+\frac{1}{x^4}={\left(11\right)}^2-2$ since $x^2+\frac{1}{x^2}=11$

$\Rightarrow x^4+\frac{1}{x^4}=121-2=119$

$\therefore x^4+\frac{1}{x^4}=119$