Question

If $x+\frac{1}{x}=9$ , then find $x^3+\frac{1}{x^3}$

• A. 729
• B. 709
• C. 702
• D. 756

Answer 1

The correct option is C

702

Given that $x+\frac{1}{x}=9$

Taking cube on both sides, we have
$(x+\frac{1}{x}{)}^3=(9{)}^3$
$\Rightarrow x^3+\frac{1}{x^3}+\left(3\right)\left(x\right)\left(\frac{1}{x}\right)(x+\frac{1}{x})=729$

$\Rightarrow x^3+\frac{1}{x^3}+\left(3\right)\left(9\right)=\left(729\right)$ ( Substituting $x+\frac{1}{x}=9$)
$\Rightarrow x^3+\frac{1}{x^3}=729-27$
$=702$