Square root of $8^3$

Question

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?

score 30 · Answer 1 · edited May 27 '13 at 00:55

30

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

edited May 27 '13 at 00:55

vadim123

82,796

answered May 27 '13 at 00:49

nitrous2

2,451

1

I like how this answer reasons according to the meaning of these ideas. – Doug Spoonwood May 27 '13 at 01:03
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 01:34
You are certainly welcome! Glad I could help. – nitrous2 May 27 '13 at 01:36

score 23 · Accepted Answer · edited May 26 '13 at 23:57

23

Using the standard rules of algebra, we compute:

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

edited May 26 '13 at 23:57

Michael Hardy

1

answered May 26 '13 at 23:55

Alex Wertheim

20,278

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

Islands · Answer 3 · 2013-05-27T01:25:30.053

8

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

edited May 27 '13 at 01:25

answered May 26 '13 at 23:56

Islands

762

Makes sense. Thank you. – user79477 May 27 '13 at 00:00
Seems a bit odd to notate index "2" in a square root. – Mario Carneiro May 27 '13 at 00:04
@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 01:24

score 0 · Answer 4 · edited Dec 25 '17 at 20:42

0

I think a key insight is the exponent rule:

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

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

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

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

$$=16\sqrt{2}$$

edited Dec 25 '17 at 20:42

Stefan4024

35,843

answered Dec 25 '17 at 19:47

guest

11

1

This adds nothing that wasn't already in other answers. – Qudit Dec 25 '17 at 20:02

Square root of $8^3$

4 Answers4