In Exercises 39–64, rationalize each denominator.5³√ -----y²

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Hello, everyone. We've been asked to eliminate the radical in the denominator by rationalizing the expression we're given is the cube root of two over K squared. We have four answer choices. All of them have no radicals in the denominator and one of them is not a fraction. So I wanna start by rewriting my original expression so that the numerator and the denominator are separate. What I mean is I recall that if I am taking a root of a fraction, I can rewrite it so that the root is on both the numerator and the denominator. So I can rewrite this as the cube root of two over the cube root of K squared. This way, I can see clear that I'm working with the denominator. Next. In order to get the radical out of the denominator, I want to multiply top and bottom of this fraction by a value that will make it. So I have a exponent that matches my index on my radical in my denominator. What I mean is we're working with a Q group. So we want the value under that radical in the denominator to have an exponent of three, right. Now we have K squared under the radical. Recalling that when I multiply bases, I add the exponents, I want to multiply top and bottom here by the cubed root of K because K multiplied by K squared will give me K cubed. So both the numerator and the denominator will be multiplied by the third root of K doing that straight across the numerator. The cube root of two multiplied by the cube root of K will result in the cube root of two K. Because since they have the same index, we can combine them under one radical in the denominator, the cu root of K squared multiplied by the cube root of K will give us the cube root of K cubed. And that's exactly what we wanted to happen. We want that index in that exponent to match. So in my next step, I'm just rewriting my num reader, it's not gonna change the cube root of two K. And in my denominator, I'm gonna simplify this. So I'm taking the cube root of K cubed that will cancel out leaving us with K in the denominator. I want to see if this can be simplified any further before I say I'm done. Since my numerator does not have a coefficient, I can't cancel the K in the denominator with anything. So this is my solution, the cube root of two K over K that is our answer and that matches answer choice. A have a nice day.

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In Exercises 39–64, rationalize each denominator.5³√ -----y²

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