###### 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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Introduction to Radicals - Problem 4

# Introduction to Radicals - Problem 3

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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Simplifying a square root of a power other than square root. So for this example what we’re actually looking at is a cube root. And the process is going to be very, very similar except for instead of needing to take out two of something, we have to take out three. So think about like square root of 4is 2, because you have 2 times 2 we have two of those. When we’re dealing with the cube rot what we actually need is 3. So the cube root of 8, is 2 times 2 times 2 is 2. You need that one more component in order to take something out of the square root.

For this problem what I’m going to do is go through and do it in one sort of long step. We’ll deal with each component, individually take out what we can and leave whatever we have to in the square root. So starting with our cube root of 64, what perfect cube goes into 64? 64 is 4 times 4 times 4 so therefore, 64 is a perfect cube, we can just take out 4 and we have nothing underneath the cube root for that particular component.

By the same logic, cube root of x³, we’re looking for things that have triplets. X³ is just x times x times x so an x comes out and we don’t have anything left in the radical.

Y to the 7th, okay, this one gets a little bit trickier. Y to the 7th is actually 7ys, okay and what we have if you think about y² times y² times y², I don’t know if you can see those, there’s a little twos I have in there. That these if we multiply our bases we add our exponents so this is actually x to the 6th. So we can take out a y² and we’re left with a y inside, the cube root. Okay, because three y²s will give us y to the 6th we have one left over.

And our last piece is this z to the 5th. Z times z times z is z³, so we can take out a z and that accounts for 3 of our 5. But we’re still left with 2 inside and there’s no way to take a cube root if we only have two things. So by going through piece by piece I was able to figure out how to simplify this cube root.

And the steps are pretty much exactly the same for a 4th root, 5th root, 6th root and so on and so forth, you just need a set of that many pieces in order to pull it out of the root.