Question

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

Answer 1

$x+\frac{1}{x}=4$

$(x+\frac{1}{x}{)}^2=16$

$x^2+\frac{1}{x^2}+2=16$

$x^2+\frac{1}{x^2}=14$

$(x^2+\frac{1}{x^2}{)}^2=196$

$x^4+\frac{1}{x^4}+2=196$

$x^4+\frac{1}{x^4}=194$