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 [tex] \sqrt[3]{2*2*2*2*3} [/tex]
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
