# How do you evaluate 2 root3 { 54} - \root[ 3] { 24} - 3\root [ 3] { 16}?

May 5, 2017

Simplify all of the radicals first, then combine like terms.

#### Explanation:

First, factor the radicands completely into prime numbers

$54 = \left(2\right) \left(3\right) \left(3\right) \left(3\right)$
$24 = \left(2\right) \left(2\right) \left(2\right) \left(3\right)$
$16 = \left(2\right) \left(2\right) \left(2\right) \left(2\right)$

Since these are cube roots, every perfect cube extracts as a single factor. So, in the case of 54, $\left(3\right) \left(3\right) \left(3\right)$ extracts as 3. The other factors remain inside their respective radicals.

$\sqrt[3]{54} = 3 \sqrt[3]{2}$
$\sqrt[3]{24} = 2 \sqrt[3]{3}$
$\sqrt[3]{16} = 2 \sqrt[3]{2}$

Our expression simplifies as

$2 \sqrt[3]{54} - \sqrt[3]{24} - 3 \sqrt[3]{16} = 6 \sqrt[3]{2} - 2 \sqrt[3]{3} - 6 \sqrt[3]{2}$

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

Observe that cube roots of 2 and cube roots of 3 would not have combined with one another. That fact is not important in this problem, though, since all of the cube roots of 2 have canceled out to zero.