Question

If $x-\frac{1}{x}=4$, Then the value of $x^4+\frac{1}{x^4}$ will be

Answer 1

Given
$x-\frac{1}{x}=4$
Squaring both side we get,
$\Rightarrow (x-\frac{1}{x}{)}^2=4^2$
$\Rightarrow x^2+(\frac{1}{x}{)}^2-2\times x\times \frac{1}{x}=4$
$\Rightarrow x^2+\frac{1}{x^2}-2=4$
$\Rightarrow x^2+\frac{1}{x^2}=6$ $......\left(1\right)$
Again squaring both side we get,
$\Rightarrow (x^2+\frac{1}{x^2}{)}^2=6^2$
$\Rightarrow x^4+\frac{1}{x^4}+2\times x^4\times \frac{1}{x^4}=36$
$\Rightarrow x^4+\frac{1}{x^4}+2=36$
$\Rightarrow x^4+\frac{1}{x^4}=34$