Question

If $x=2^{\frac{1}{3}}-2^{\frac{-1}{3}}$, then $2x^3+6x$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. None of these

Answer 1

The correct option is D None of these

We are given,

$x=2^{\frac{1}{3}}+2^{-\frac{1}{3}}$

$2x^3-6x=2{(2^{\frac{1}{3}}+2^{-\frac{1}{3}})}^3-6x$

$=2[{\left(2^{\frac{1}{3}}\right)}^3+{\left(2^{-\frac{1}{3}}\right)}^3+3{\left(2^{\frac{1}{3}}\right)}^2\left(2^{\frac{-1}{3}}\right)+3\left(2^{\frac{1}{3}}\right){\left(2^{\frac{-1}{3}}\right)}^2]-6x$

$=2(2+2^{-1}+3\left(2^{\frac{2}{3}-\frac{1}{3}}\right)+3\left(2^{\frac{1}{3}-\frac{2}{3}}\right))-6x$

$=2(2+\frac{1}{2}+3(2^{\frac{1}{3}}+2^{\frac{-1}{3}}))-6x$

$=4+1+6(2^{\frac{1}{3}}+2^{\frac{-1}{3}})-6x$

$=5+6x-6x$

$=5$

Hence, this is the answer.

Option $\left(D\right)$ is correct.