Question

Simplifying Nonperfect Roots

Answer 1

Short Answer:Choice C [Third one down]
Remark
Start with the number so I can talk about the basic principle at work. The way to work it is to factor 162 into prime factors and hope there are at least 3 that are the same. 
162 = 2 * 81
162 = 2 * 3 * 3 * 3 * 3

Now When you take the cube root of that, you get $\sqrt[3]{2*2*2*2*3}$
Here's the rule for a cube root. For every 3 prime factors under the cuberoot sign, you take out one and throw the other two away. So for cube root of 81,
you would factor it as ∛(81) = ∛(3 * 3 * 3 * 3) = 3∛3.

So out of four 3s under the cube root sign,  you have 1 outside the root sign and one inside the root sign. 2 of the four 3s have been thrown away.

Continuation.
X first
You want 2 xs outside the root sign
∛(x * x * x * x * x *x ) = (x * x)  You have thrown away 2 xs for every x outside the cube root sign
C = 6 There are no left overs.

Y second
For the ys, you need 1 outside and 2 inside the root sign. that's because you need 5 altogether.
∛(y * y * y * y * y) = y ∛y^2

Answer 
c = 6; d = 2
Choice C <<<<<< answer