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

Algebra.Com
Question 460307:

Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!

=
=
=
=

Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!

Multiply under the radicals and get



Break it into the quotient of two fifth roots:



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 




We multiply under the radicals:



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




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




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



Edwin

RELATED QUESTIONS

{{{5th root(8x^(3)/y^(4))* 5th... (answered by ankor@dixie-net.com)
3cubed root y^4 * 3 cubed root 16 y^5 how do you do this... (answered by Gogonati)
{{{fthrt(8x^3/y^4) *... (answered by Alan3354)
root 5 + 2* root 3 mutiply with 7 + 4*root... (answered by tommyt3rd)
square root of y+1 plus square root of... (answered by edjones)
Pascals Triangle :Please i need help with this thanks solve: 1)(3a^2+b)^4... (answered by Edwin Parker)
Please simplify and explain: Cubed root of (2x) • cubed root of (4x^2•y^2) • cubed root... (answered by Alan3354)
I need some help I am not sure what method was used to solve this problem... (answered by funmath,stanbon)
Eliminate the parameter. x = t + 4, y = t2 (5 points) choices below y = square... (answered by MathLover1)