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
