Question

If $(x+\frac{1}{x})=2\sqrt{3}$, then the value of $(x^3+\frac{1}{x^3})$ is

• A. $12\sqrt{3}$
• B. $18$
• C. $18\sqrt{3}$
• D. None of these

Answer 1

The correct option is C $18\sqrt{3}$

Given, $(x+\frac{1}{x})=2\sqrt{3}$

By cubing on both sides, we get

${(x+\frac{1}{x})}^3={\left(2\sqrt{3}\right)}^3$

$\Rightarrow x^3+\frac{1}{x^3}+3x^2\cdot \frac{1}{x}+3x\cdot \frac{1}{x^2}=24\sqrt{3}$ ....Since $({(a+b)}^3=a^3+b^3+3a^{2}b+3a{b}^2)$

$\Rightarrow x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=24\sqrt{3}$

$\Rightarrow x^3+\frac{1}{x^3}=24\sqrt{3}-3\times 2\sqrt{3}$

Therefore, $(x^3+\frac{1}{x^3})=18\sqrt{3}$