Question

If $p=x^{\frac{1}{3}}+x^{\frac{-1}{3}}$, then $p^3-3p$ is equal to

• A. $-2$
• B. $\frac{1}{2}(x+x^{-1})$
• C. $x+x^{-1}$
• D. $2(x+x^{-1})$

Answer 1

The correct option is C $x+x^{-1}$

Given,

$\Rightarrow p=x^{\frac{1}{3}}+x^{\frac{-1}{3}}$

$\therefore p=x^{\left(\frac{1}{3}\right)}+\frac{1}{x^{\frac{1}{3}}}$ $...\left(i\right)$.

We know, $(a+b{)}^3=a^3+b^3+3ab(a+b)$.

Taking cube on both sides of $\left(i\right)$, we get,

$\Rightarrow p^3=x^{(\frac{1}{3}\times 3)}+(\frac{1}{x}{)}^{(\frac{1}{3}\times 3)}+3\times 1\times (x^{\frac{1}{3}}+x^{\frac{-1}{3}})$

$\Rightarrow p^3=x+\frac{1}{x}+3(x^{\frac{1}{3}}+x^{\frac{-1}{3}})$

$\Rightarrow p^3-3p=x+\frac{1}{x}=x+x^{-1}$.

Hence, option $C$ is correct.