Question

If $(x-\frac{1}{x})=6,$ find the value of $(x^2+\frac{1}{x^2})$ and $(x^4+\frac{1}{x^4})$

Answer 1

$x-\frac{1}{x}=6$;

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

$\Rightarrow 36=x^2+\frac{1}{x^2}-2$

$\Rightarrow x^2+\frac{1}{x^2}=38$

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

$(38{)}^2=x^4+\frac{1}{x^4}+2$

$\Rightarrow x^4+\frac{1}{x^4}=1442$.