How do I simplify this Algebra Problem?

Question

the 4th root (like cube root) of 162 c^2 d^6

Answer 1

(162 c^2 d^6)^(1/4) =
(2 * 3^4 c^2 d^6)^(1/4) =
3d (2 c^2 d^2)^(1/4)

You can simply further, because of the even powers of c and d (the fourth root of c^2 is the same as the square root of |c|)...

3d (2 c^2 d^2)^(1/4) =
3d sqrt(|cd|) 2^(1/4)

... but you can't get rid of the fourth root of 2.

Answer 2

4th root of (162 c^2 d^6)

prime factor

4th root of (2 x 3 x 3 x 3 x 3 c^2 d^4 d^2)

4th root of (2 x 3^4 c^2 d^4 d^2)

3d 4th root of (2 c^2 d^2) => this answer is valid if the variables are positive. If not then the answer is 3 l d l 4th root of (2 c^2 d^2)

Answer 3

Alright, first make it the fourth root of (81*2) c^2 d^4 d^2

81=3^4, so you can take the 3 and the d^4 out, giving you
3d on the fourth root of 2 c^2 d^2

Answer 4

This is basically like the other root problems. So when you find the prime factorization, what you will have under the root is
3*3*3*3*2*c*c*d*d*d*d*d*d
In the forth root you need four of each number in order to bring one out. You have four 3's and 4 d's. This leaves you one 2, two c's and two d's. Since you don't have four of them, they need to stay under the root.

After bringing out a 3 and a d, you get your answer:
3d forth roots of 2c^2d^2