# Square root of $8^3$

I'm only in intermediate algebra. I know that $\sqrt{8^3}$ is equal to $16\sqrt{2}$ but could you simply explain the process on how to get to that?

We can reduce $8^3$ to its prime factors: $$8^3=512\implies 2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$$ Because this is a square root, we are looking for groups of $2$. Similarly, if this was a cube root we would look for groups of 3: $$\boxed{2\cdot2}\boxed{2\cdot2}\boxed{2\cdot2}\boxed{2\cdot2}2$$ We have 4 groups of 2 which we will take out of the radical: $$2\cdot2\cdot2\cdot2\sqrt{2}$$ We can now simplify this as: $$\boxed{16\sqrt{2}}$$

• I like how this answer reasons according to the meaning of these ideas. May 27, 2013 at 1:03
• Thank you so much for contributing! This is actually how I learned it in the beginning. Reduce the numbers as far as you can, pair them up, turn the pairs into single numbers, multiply those, and the remainder that didn't pair goes in the radical. May 27, 2013 at 1:34
• You are certainly welcome! Glad I could help. May 27, 2013 at 1:36

Using the standard rules of algebra, we compute:

$$\sqrt{8^{3}} = \sqrt{8^2 \cdot8} = \sqrt{8^2}\cdot\sqrt{8} = 8\cdot\sqrt{4\cdot2} = 8\cdot\sqrt{2^2}\cdot\sqrt{2} = 16\sqrt{2}$$

• Very simple. Thanks! May 26, 2013 at 23:59
• No problem, glad I could help! May 26, 2013 at 23:59

$\sqrt[]{8^3}$ $= \sqrt[]{8^2\cdot8}$ $= 8\sqrt[]{8}$ $=8\sqrt[]{2\cdot4}$ $=2\cdot8\sqrt[]{2}$ $=16\sqrt[]{2}$

• Makes sense. Thank you. May 27, 2013 at 0:00
• Seems a bit odd to notate index "2" in a square root. May 27, 2013 at 0:04
• @Mario Carneiro: It does. I'm still getting used to latex on here and it was just a habit. Fixed it. May 27, 2013 at 1:24

I think a key insight is the exponent rule:

$$\sqrt {8^3}=$$

$$\sqrt {(2^3)^3}=$$

$$\sqrt {2^9}=$$

$${2^4} \cdot \sqrt{2}$$

$$=16\sqrt{2}$$

• This adds nothing that wasn't already in other answers. Dec 25, 2017 at 20:02