# How do you simplify  sqrt(27k^7q^8)?

There are different ways to look at how to simplify root statements - I present one of mine below and get to $3 {k}^{3} {q}^{4} \sqrt[2]{3 k}$

#### Explanation:

When we're dealing with simplifying a term within a root, I like to use the "buddy" rule - which is, in order for a term to "escape" the root, it needs the number of buddies listed in the root (a square root is 2 buddies, a cube root is 3 buddies, etc). When there is no number, like in this case, the default is 2.

Let's start with the original (but I'm going to add the 2 to indicate clearly this is a square root):

$\sqrt[2]{27 {k}^{7} {q}^{8}}$

Now let's break down each part of the terms inside the root piece by piece:

27

$27 = \left(3 \cdot 3\right) \cdot 3$, which means there is one of 3 "buddies" and one last 3. So one of the 3 "buddies" gets out and the last 3 stays inside the root.

${k}^{7}$

${k}^{7} = \left(k \cdot k\right) \cdot \left(k \cdot k\right) \cdot \left(k \cdot k\right) \cdot k$, which means there are 3 sets of k "buddies" so 3 k's get out from under the root and the last k is left inside the root.

${q}^{8}$

${q}^{8} = \left(q \cdot q\right) \cdot \left(q \cdot q\right) \cdot \left(q \cdot q\right) \cdot \left(q \cdot q\right)$, which means there are 4 sets of q "buddies" so 4 q's escape the root and none stay inside the root.

Putting it all together, we get:

$3 {k}^{3} {q}^{4} \sqrt[2]{3 k}$