Question

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

Answer 1

$x+\frac{1}{x}=3$

${(x+\frac{1}{x})}^2=(3{)}^2=9$

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

$x^2+{\frac{1}{x}}^2=9-2=7$

${(x^2+{\frac{1}{x}}^2)}^2=(7{)}^2=49$

$x^4+{\frac{1}{x}}^4+2=49$

$x^4+{\frac{1}{x}}^4=49-2=47$