Question

If $x+\frac{1}{x}=4$, then the value of $x^3+\frac{1}{x^3}$ is:

• A. $52$
• B. $64$
• C. $68$
• D. $76$

Answer 1

The correct option is A $52$

Given, $x+\frac{1}{x}=4$.

On taking cubes on both sides,
$\therefore (x+\frac{1}{x}{)}^3=4^3$
$\therefore x^3+\frac{1}{x^3}+3.x.\frac{1}{x}(x+\frac{1}{x})=64$
$\therefore x^3+\frac{1}{x^3}=64-3.x.\frac{1}{x}(x+\frac{1}{x})$

$\therefore x^3+\frac{1}{x^3}=64-3(x+\frac{1}{x})$
$=64-3\left(4\right)$
$=64-12=52$.

Option $A$ is correct.