# How do you solve root3(2x^3 - 24) = -x and find any extraneous solutions?

Jun 27, 2016

Cube both sides.

#### Explanation:

${\left(\sqrt[3]{2 {x}^{3} - 24}\right)}^{3} = {\left(- x\right)}^{3}$

$2 {x}^{3} - 24 = - {x}^{3}$

$3 {x}^{3} = 24$

${x}^{3} = 8$

$x = 2$

Checking in the original equation:

root(3)(2(2)^3 - 24 )=^? -2

$\sqrt[3]{- 8} = - 2$

Therefore, out solution set is $\left\{x = 2\right\}$.

Hopefully this helps!