Question

If $x^2+\frac{1}{x^2}=83$. Find the value of $x^3-\frac{1}{x^3}$.

Answer 1

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

${(x-\frac{1}{x})}^2=83-2$

${(x-\frac{1}{x})}^2=81$

$x-\frac{1}{x}=9$

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

$x^3-\frac{1}{x^3}-3(x-\frac{1}{x})=729$

$x^3-\frac{1}{x^3}-3\times 9=729$

$x^3-\frac{1}{x^3}=729+27=756$