# Question #7c61d

Jan 28, 2017

$x > - 1$

#### Explanation:

${x}^{3} > - 1$

Cube root both sides:

$\sqrt[3]{{x}^{3}} > \sqrt[3]{- 1}$

The cuberoot of x cubed is just x, and the cube root of -1 is -1.
This is because: $\left(- 1\right) \times \left(- 1\right) \times \left(- 1\right) = - 1$

So: $x > - 1$

Hope this helps!