Radicals and Absolute Values - Problem 3 2,900 views
Simplifying square roots or cube roots keeping absolute value in mind. So what we're going to be looking at now is the cube root of a big string of information. And what we need to remember about cube roots is that we can put in positive or negative numbers and positive or negative numbers can come out.
So I think about the cube root of 8 is 2 and the cube root of -8 is -2, so either thing can go in either thing can come out. Alright, so looking at this problem what we are looking at is just four components and I’m going to do them all at once walking you through my thought process.
So starting through with the cube root of -64. 4 to the third is 64, dealing with the negative, -4 to the third is -64. So the cube root of -64 is just going to be -4. Negative numbers can come out of odd roots so we are okay we don’t need our absolute values.
Moving down the row, the cube root of x³. Very similar to this we just have something to the third. The cube root of something to the third is just that component, this is x. And x can be positive or negative because it's coming out of an odd root, if it was an even root we would have to consider our absolute values, but we are not. Cube root of y to the 7th, y to the 6th is y² to the third so this is just going to be y² and we are still left with a cube root of y left over.
And our last on is the cube root of z to the 5th. Z³ is obviously 3z’s so I can take out one of those z’s and we are still left with two that are stuck behind.
So broke it down bit by bit and whenever we are dealing with an odd root, we don’t have to think about absolute values because positive or negative numbers can come out of them. We only need to think about absolute values when we are dealing with an even root or square root of 4th root so on and so forth.