# How do you write the simplified form of -64^(1/ 3)?

Apr 24, 2018

the simplified answer would be -4

#### Explanation:

Let's factor out 64:
$64 = {2}^{6}$
$- {\left({2}^{6}\right)}^{\frac{1}{3}}$
$= - {2}^{6. \left(\frac{1}{3}\right)}$
$= - {2}^{2}$
$= - 4$

Apr 24, 2018

$- 4$

#### Explanation:

Recall one of the laws of indices:

$\sqrt{x} = {x}^{\frac{1}{2}} \text{ "and " } \sqrt[3]{x} = {x}^{\frac{1}{3}}$

$- {64}^{\frac{1}{3}} = \sqrt[3]{- 64}$

$64$ is a perfect cube: $64 = {4}^{3}$

$\sqrt[3]{- 64} = - 4$

You could also work with the prime factors:

$\sqrt[3]{- 64} = \sqrt[3]{- \left({2}^{6}\right)}$

$= - {2}^{2}$

$= - 4$

Note that perfect cubes can be negative, but perfect squares cannot.