# 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?

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Using the standard rules of algebra, we compute:

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

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Very simple. Thanks! – user79477 May 26 '13 at 23:59
No problem, glad I could help! – Alex Wertheim May 26 '13 at 23:59

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: $$\boxed{16\sqrt{2}}$$

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I like how this answer reasons according to the meaning of these ideas. – Doug Spoonwood May 27 '13 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. – user79477 May 27 '13 at 1:34
You are certainly welcome! Glad I could help. – nitrous2 May 27 '13 at 1:36

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

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Makes sense. Thank you. – user79477 May 27 '13 at 0:00
Seems a bit odd to notate index "2" in a square root. – Mario Carneiro May 27 '13 at 0:04
@Mario Carneiro: It does. I'm still getting used to latex on here and it was just a habit. Fixed it. – Islands May 27 '13 at 1:24