SOLUTION: {{{root(5, (8x^3)/(y^4))*root(5, (4x^4)/(y^2))}}}

Algebra ->  Radicals -> SOLUTION: {{{root(5, (8x^3)/(y^4))*root(5, (4x^4)/(y^2))}}}       Log On


   

Question 460307: root%285%2C+%288x%5E3%29%2F%28y%5E4%29%29%2Aroot%285%2C+%284x%5E4%29%2F%28y%5E2%29%29
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
root%285%2C+%288x%5E3%29%2F%28y%5E4%29%29%2Aroot%285%2C+%284x%5E4%29%2F%28y%5E2%29%29
Multiply under the radicals and get

root%285%2C+%2832x%5E7%29%2F%28y%5E6%29%29

Break it into the quotient of two fifth roots:

root%285%2C+32x%5E7%29%2Froot%285%2Cy%5E6%29

The power of y in the denominator must be a
multiple of 5, so it needs to be multiplied by
y4 so that it will become y10.

So we multiply the entire fraction by root%285%2Cy%5E4%29%2Froot%285%2Cy%5E4%29




We multiply under the radicals:

expr%28root%285%2C+32x%5E7y%5E4%29%2Froot%285%2Cy%5E10%29%29

The denominator is simplified by dividing the exponent 10
by the index of the root 5, getting exponent 2, and 
eliminating the radical,  so we have

expr%28root%285%2C+32x%5E7y%5E4%29%2Fy%5E2%29


Now we will simplify the numerator.
Write 32 as 25
Write x7 as x5x2


expr%28root%285%2C+2%5E5x%5E5x%5E2y%5E4%29%2Fy%5E2%29

Finally you can take the two fifth powers out
of the fifth root and have 2x in front of the
radical:

%282x%2Aroot%285%2C+x%5E2y%5E4%29%29%2Fy%5E2

Edwin