Question

If $x^2+\frac{1}{x^2}=98$, then find the value of $x^3+\frac{1}{x^3}$

Answer 1

$x^2+\frac{1}{x^2}=98$

${(x+\frac{1}{x})}^2-2\times x\times \frac{1}{x}=98$

$\Rightarrow {(x+\frac{1}{x})}^2-2=98$

$\Rightarrow {(x+\frac{1}{x})}^2=98+2=100$

$\Rightarrow x+\frac{1}{x}=\pm 10$

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

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

$=(\pm 10)(98-1)$

$=\pm 970$