Question

Am I right?

Answer 1

MAYBE.

Answer 2

\[\sqrt[3]{32}\]

This is the answer I ended with.
\[8\sqrt{8}\]

Answer 3

I think it is wrong, because I'm not sure what I'm supposed to do with the exponent :c

Answer 4

Do you have answer choices?

Answer 5

No, all I know is that I need to simplify the equation and the answer cannot be a decimal under any circumstances.

Answer 6

No, that's not right. Remember that the perfect cubes are
\[2^3 = 8\]
\[3^3 = 27\]
etc. So
\[\sqrt[3]{32} = \sqrt[3]{ 8 \cdot 4} = \sqrt[3]{8} \cdot \sqrt[3]4 = ?\]

Answer 7

So whenever there is an exponent on the outside I'm supposed to multiply the simplified version of 32?

Answer 8

Okay--- You lost me Jem. I know that, you can only simplify 4 into a perfect square which would be 2... But would it be, 4^3 or not?

Answer 9

You can't simplify the cube root of four, but you can certainly simplify the cube root of eight.

Answer 10

Wait what--- 8 isn't a perfect square though? How is that possible?

Answer 11

\(\sqrt[3]{32} \)

\( = \sqrt[3] {2 \times 2 \times 2 \times 2 \times 2 } \)

\( = \sqrt[3] {\color{red}{2 \times 2 \times 2}}~ \sqrt[3]{ \color{green}{2 \times 2} } \)

\( = \color{red}2 ~ \sqrt[3]{ \color{green}4 } \)

Answer 12

You're not talking about perfect squares, you're talking about perfect cubes.

Answer 13

8 is a perfect cube which is what you need here since you are taking the cubic root.

Answer 14

I never learned about that, so I don't know how to do it...

Answer 15

It's exactly the same idea as a square root.
\[2^2 = 2\cdot 2 = 4 \implies \sqrt{4} = 2\]
\[2^3 = 2\cdot 2 \cdot 2 = 8 \implies \sqrt[3]8 = 2 \]

Answer 16

So the answer is just 2?

Answer 17

Answer is

\(2\sqrt[3]{4} \)

Answer 18

I wish I could give a medal to the both of you-- But, Jem helped first.. Sorrry Math

Answer 19

@TheHero
You are familiar with square roots.
Can you simplify \(\sqrt{27}\) showing steps?

Answer 20

Yes, that would be easy.

\[\sqrt{27}\]

\[\sqrt{3} \sqrt{9} = 3\sqrt{3}\]

Answer 21

Thanks, but don't worry about the medal. The important thing is for you to understand this.

Answer 22

Great. The reason you chose to factor 27 into 9 and 3 is that you are taking a square root, and 9 is a perfect square.

When you take a cubic root, then you factor the number trying to take out the largest perfect cube.

The number 32 = 2 * 2 * 2 * 2 * 2.

If you take three factors of 2, 2 * 2 * 2 = 8, you have a perfect cube.
Thats' why the number 32 was factored into 8 and 4.
Since 8 is a perfect cube, the cubic root of 8 is 2.
4 is not a perfcet cube, so it remained inside the cubic root sign.

This is similar to the problem \(\sqrt{27}\)you did correctly, where the 3 remained inside the square root and the \( \sqrt{9} \) came out as 3.

Answer 23

Oh! I understand it now, thank you very much! c:

Answer 24

For example,

\( \sqrt[5] {729} \)

\(= \sqrt[5] {3^6} \)

\( =\sqrt[5]{3^5} ~ \sqrt[5] {3} \)

\(=3 \sqrt[5]{3} \)

Since this problem has a fifth root, you need to take out a perfect fith power number, 3^5.

Answer 25

You're welcome.