Question

If $x=\sqrt[3]{32}-\sqrt[3]{3\frac{1}{2}}$ , then $2x^3+6x=$

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

Answer 1

The correct option is C $3$

Given ,

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

which imply

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

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

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

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

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

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

$2x^3=$ $3-6x$

$2x^3+6x=$ $3$