SOLUTION: cube root (x^2y^4) x cube root (x^4y^10) i got : x^2 cube root(y^10) is this correct

Algebra ->  Square-cubic-other-roots -> SOLUTION: cube root (x^2y^4) x cube root (x^4y^10) i got : x^2 cube root(y^10) is this correct      Log On


   

Question 408332: cube root (x^2y^4) x cube root (x^4y^10)
i got : x^2 cube root(y^10) is this correct
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Please don't use "x" for multiplication. Use the word "times" or use "*" (Shift+8).

root%283%2C+x%5E2y%5E4%29+%2A+root%283%2C+x%5E4y%5E10%29
Your answer, x%5E2%2Aroot%283%2C+y%5E10%29 is correct so far but it is unfinished. There is more simplifying we can do with the remaining cube root.

y%5E10 is not a perfect cube. (Variables are perfect cubes if their exponents are multiples of 3.) But it does have perfect cube factors. Factoring the y%5E10 into as many perfect cube factors as possible we get:
x%5E2%2Aroot%283%2C+y%5E3%2Ay%5E3%2Ay%5E3%2Ay%29
Now we can use a property of radicals, root%28a%2C+p%2Aq%29+=+root%28a%2C+p%29%2Aroot%28a%2C+q%29 to separate the factors inside the cube root into separate cube roots:

The cube roots of the perfect cubes will simplify:
x%5E2%2Ay%2Ay%2Ay%2Aroot%283%2C+y%29
which simplifies to:
x%5E2%2Ay%5E3%2Aroot%283%2C+y%29
This is the fully simplified expression.