Question

I need help finding the answers to ^3sqrt-343 and ^3sqrt64. Can you show me the steps so I can learn how to do them.

Answer 1

Just to clarify, this is how it is written with those ^3 ?
\( \sqrt[3]{343} \text{ and } \sqrt[3]{64} \)

Answer 2

It usually comes down to testing multiplicities. You want to find the prime factorization of each number. Anything with a power 3 or a multiple of 3 can come through the radical by dividing the power by 3 once it comes outside.

Answer 3

Yes. Plz.

Answer 4

So, I would start with the radical with 64. That one is easier!
Your goal is essentially prime factorization of each number. Powers of 3 or multiples are able to be simplified in the cube root.
This one is a common square to memorize. If you know 8* 8 = 64, you're already getting somewhere.
\( \sqrt[3]{64} = \sqrt[3]{8 * 8} \)
You dan then divide each 8 in half for 4. The 2's are taken independently.
\( \sqrt[3]{2*4 * 2*4} \)
Now 4 may be divided in half.
\( \sqrt[3]{2*(2* 2) * 2 * (2* 2)} \)
This can be written as a power because we have six 2's multiplied together. 2*2*2*2*2*2 = 2^6
\( \sqrt[3] {2^6} \)
We can take this out of the radical by dividing its power by 3.
\( 2^{6/3} \sqrt[3]{1} \)
It is trivial to write this step. The cube root of 1 is simply 1, so it disappears. But it is good to know the radical was there when I removed the 2^6. Simplify the power and you get...
\(2^2 = 4 \)

Answer 5

343 is more tricky because you probably do not see this one often. It actually is a perfect cube as well, but you'd have to exhaust your options first to see it.
You could go through this sort of elimination process.
- Is it divisible by 2? [No, 343 is odd!]
- Is it divisible by 3? [No, 3+4+3 = 10 is not divisible by 3]
- Is it divisible by 4? 5? 6? [No; No; and No; Last digit is not even, nor 0 or 5, nor totally divisible by 2 and 3]

Answer 6

You would get to 7, which is more unusual to test. I suggest just trying to divide that one off in long division.
4 9
7 /343
28
63
63
0
343 is divisible by 7. Neat! And notice its dividend 49 is also 7^2.
We end up here, rewriting 343: \( \sqrt[3]{7*7^2} = \sqrt[3]{7^3} \)

Answer 7

So to summarize, if you know your multiplicity rules / how to prime factorize numbers, that is often the first step. The prime factorization often has powers of primes that can be reduced when taken out of the cube root. The whole process takes some patience and trial yet after some practice, it can become second nature!

If I need to clarify anything, I am happy to. :)

Answer 8

I don't understand the last few step in getting 4.

Answer 9

\( \sqrt[3]{2^6} \)
From this point?

Answer 10

Yep.

Answer 11

There are a few ways to really reduce this logically. The rules of powers could become useful in justifying the step I made.
\( \sqrt[3]{2^6} \)
We could express this in two ways that make it easy to see the result. One way is the power to a power. 6 = 2*3, so we can rewrite this with two powers:
\( \sqrt[3]{(2^2)^3} \)
The cube and cube root are inverse operations, so they cancel. Does that seem better?

Answer 12

The other way is to rewrite it as a product of two 2^3. We can separate it using properties of radicals instead.
\( \sqrt[3]{2^6} = \sqrt[3]{2^3 \times 2^3} \)
\( = \sqrt[3]{2^3} \times \sqrt[3]{2^3} \)
As you can see, in each case you get 4.

Answer 13

Ok. I get it now. Thanks you for helping me understand. You've helped a lot.

Answer 14

Glad to be able to help! :)