To do this problem, we're going to start by taking that 32 in prime factoring it. So 32 is 4 * 8, and four simplifies to 2 * 2, and eight is 4 * 2. And so if we keep breaking that down, we get 2 to the fifth being the same thing as 32. Now a really cool method is to think about this five goes in to the exponent 5 how many times? And the answer is one complete time. So 2 to the first five goes into five one whole time. Then we could think about that five goes into this 11. How many times the five goes into 11? Twice. But there's some leftovers now, so we're going to put the leftovers still inside the radical. SO2 X squared. The 5th root of X is our final answer. There's some other ways we could look at this. We could think about doing 2 to the 5 / 5, because this index is really just like being in the denominator of a fraction. So X to the 11 / 5 5 / 5 we can see is to the first power. Anything to the first power is really itself. This X to 11 fifths we could think of as 10 fifths plus one more fifth. Well, 10 fifths is really two whole and one fifth. Now when we have multiplication, when we have addition and exponents, it's the same thing as multiplication in the bases. So I could think of X ^2 X 2 + 1/5 as X ^2 * X to the one fifth that X ^2 would be a whole quantity and then that 1/5 portion we could put back into the 5th root of X. So more than one way to do it. But the shortcut method is just think about the index goes into the exponent. How many full times? Full times go on the outside, the leftovers go on the inside. When doing this problem, the first thing I would recommend to do is to factor the 216. So 216. Obviously A2 can come out. So 2 and 108 and another 2SO2 and 54 and then another oh 54 we could think of as maybe 6 * 9. So keep bringing these twos down. Six 2 * 3 and nine 3 * 3. So if we look at this, the prime factorization of 216 is 2 ^3 and three cubed and then X to the 13th. So what we're going to do is we're going to think about that index of three right here goes into each of those exponents. How many times? Well, three goes into three. I'm going to rewrite it. It got messy. Three, 2 ^3, 3 ^3, and X to the 13. So two goes into the exponent of three, three goes into the exponent of three. How many times? And the answer is once. The same index of three goes into the next exponent of three how many times? Once with no leftovers, now three goes into the exponent of 13. How many times? And the answer is 4, but there's a leftover. The leftovers are always going to stay within the box, so it's going to be 2 * 3 or 6X to the 4th, the cube root of X.