Question

If $x+\frac{1}{x}=11$ , then find $x^3+\frac{1}{x^3}$

• A. 1364
• B. 1298
• C. 1331
• D. 121

Answer 1

The correct option is B

1298

Given that $x+\frac{1}{x}=11$

Taking cube on both sides, we have
$(x+\frac{1}{x}{)}^3=(11{)}^3$
$\Rightarrow x^3+\frac{1}{x^3}+\left(3\right)\left(x\right)\left(\frac{1}{x}\right)(x+\frac{1}{x})=1331$

$\Rightarrow x^3+\frac{1}{x^3}+\left(3\right)\left(11\right)=\left(1331\right)$ ( Substituting $x+\frac{1}{x}=11$)
$\Rightarrow x^3+\frac{1}{x^3}=1331-33$
$=1298$