SOLUTION: cube root (8m^7n^9/n^2m^2) i got: cube root 8m^5n^7 8 ^ 2 *4 ^ 2*2 final answer: 2cube root m^5n^7 ???? just want to make sure I did it right.

Algebra ->  Radicals -> SOLUTION: cube root (8m^7n^9/n^2m^2) i got: cube root 8m^5n^7 8 ^ 2 *4 ^ 2*2 final answer: 2cube root m^5n^7 ???? just want to make sure I did it right.      Log On


   

Question 405428: cube root (8m^7n^9/n^2m^2)
i got:
cube root 8m^5n^7
8
^
2 *4
^
2*2
final answer: 2cube root m^5n^7 ????
just want to make sure I did it right.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
root%283%2C+%288m%5E7n%5E9%29%2F%28n%5E2m%5E2%29%29
Your first step is right on. The radicand (the expression inside the radical) simplifies to:
root%283%2C+8m%5E5n%5E7%29
Next we look for factors of the radicand that are perfect cubes. As you already found, 8 is a perfect cube. But there are more perfect cube factors. Because of the way exponents work, the exponent on a perfect cube is not a perfect cube but a multiple of 3! So x%5E3, y%5E12, z%5E300 are all perfect cubes, even though 3, 12 and 300 are not themselves perfect cubes.

So your radicand factored into as many perfect cubes as we can find is:
root%283%2C+8m%5E3%2Am%5E2%2An%5E3%2An%5E3%2An%29
For reasons that will become clear shortly I like to use the Commutative Property to rearrange the order of the factors so that all the perfect cubes are in front:
root%283%2C+8m%5E3%2An%5E3%2An%5E3%2Am%5E2%2An%29
Next we use a property of radicals, root%28a%2C+p%2Aq%29+=+root%28a%2C+p%29%2Aroot%283%2C+q%29, to split this cube root of a product into a product of cube roots. We want each perfect cube factor in its own cube root. The factors that are not perfect cubes can all go into one cube root:

The cube roots of the perfect cubes will simplify:
2%2Am%2An%2An%2Aroot%283%2C+m%5E2%2An%29
or
2%2Am%2An%5E2%2Aroot%283%2C+m%5E2%2An%29
This is the simplified cube root. (Note how the radical is at the end. This is the usual way to write terms like this and it is the reason I put all the perfect cubes n the front earlier.)