###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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# Rationalizing the Denominator with Higher Roots - Problem 2

Carl Horowitz
###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Rationalizing the denominator of an expression of higher power, when we have more than one element in that radical. So the first step, whenever I am trying to rationalize a denominator, is to simplify that root if I can at all.

So looking at this, I have the cube root of three components. The first thing I want to do is just simplify this, is there anything I can take out of that root to make my life easier?

16 is 2 times 8 and I know 8 is a cubic of 2. So we can take out that leaving us with 2 and we're left with the cube root of 2 inside. The cube root of x to the fourth, that's just going to be x with a x left in the radical. And the cube root of y to the eighth, yÂ² with 2y's left over and we still have our numerator to be 7.

I don't need any absolute values in this case, because we're dealing with a odd root, so therefore everything can be positive or negative, it doesn't really matter.

So what I've done here is I've taken my root in the denominator and basically simplified it to make my numbers smaller. And now we need to rationalize this. We need to somehow get rid of this cube root in the denominator. In order to do that whenever we're dealing with a cube root in the denominator, we're going to need 3 of these component in order to get rid of that root.

So what we need to do is multiply by a cube root. We need to get rid of the 2. In order to get rid of the 2, I need 3 of them because we're in the cube root. So I need to multiply by 2Â². You could write 4 if you want, I tend to like leaving things with power so I can see the breakdown of here is 1 and here is 2. I have a single x, I need 3xs, so I need 2 more xs. And I have yÂ², I need 3 of them, so I only need one more y. Whatever we multiply by in the denominator, we need to multiply by in the numerator. This is 4xÂ² y and then simplify.

So nothing we can combine in the numerator, so we're just left with 7 cube root of 4xÂ² y, the denominator 2x yÂ² from before and then we're left with these two radicals multiplied together.

The way that we choose the square root, this radical rather is to have three of each of these components, so we multiply them together, our radical goes away and we're just left with 2xy. So this term came from multiplying these two roots together, simplify the denominator up, we're left with 7 oops that's right where I have a little bit more room let's go over this way. So we're left with 7 cube root of 4xÂ² y. Combining like terms 2 times 2 gives us 4, x times x gives us xÂ², y times yÂ² gives us y cubed.

It's pretty involved to getting rid of your radical from the denominator, but all steps are necessary. So the first thing we did was to simplify that radical, make our numbers smaller, then we had to multiply by something in order to get rid of those terms in that radical and then simplify it up.