Is My Solution to the Complex Algebraic Expression Correct?

[mathdrama]

I had to evaluate this: [32x^5y^4)^2/5/(4x^2)^2y^3/5]^-1

This is how I did it, but is it correct?

= [32^2/5x^2y^4(2/5)-1
(4x^2)^(2y^3/5)

= [32^2/5x^2y^8/5)^-1
(4x2)^(2y^3/5)

= [(^5√32)^2*x^2y^8/5)-1
(4x2)^(2y^3/5)

= [(2)^2*x^2y^8/5)-1
(4x^2)^(2y^3/5)

= [4*x^2y^8/5]^-1
(4x^2)^(2y^3/5)= [4x^2y^8/5]-1
(4x^2)^(2y^3/5)= (4x^2)(^2y^3/5)
4x^2y^8/5
I apologize for all the messiness.

 

[Opalg]

Apart from all the messiness (of which there is a lot), I think that you may be getting towards the right answer. The final line of the calculation, $\dfrac{\text{(4x^2)(^2y^3/5)}}{\text{4x^2y^8/5}}$, makes no sense at all as it stands. But if you put one of the parentheses in the top row in a different position, so that it reads $\dfrac{\text{(4x^2)^2 y^(3/5)}} {\text{4x^2y^8/5}} = \dfrac{(4x^2)^2 y^{3/5}}{4x^2y^{8/5}}$, then you are on the right lines. But you can still simplify this further, by cancelling some of the stuff in the numerator with parts of the denominator.

 

[soroban]

Hello, mathdrama!

I think your problem looks like this:

\text{Simplify: }\:\left[\frac{(32x^5y^4)^{\frac{2}{5}}}{(4x^2)^2y^{\frac{3}{5}}}\right]^{-1}

A fraction to the minus-one power is inverted: .\left(\tfrac{a}{b}\right)^{-1} \:=\:\tfrac{b}{a}

\left[\frac{(32x^5y^4)^{\frac{2}{5}}}{(4x^2)^2y^{\frac{3}{5}}}\right]^{-1}

. . =\;\frac{(2^2\cdot x^2)^2\cdot y^{\frac{3}{5}}}{(2^5\cdot x^5\cdot y^4)^{\frac{2}{5}}}

. . =\;\frac{(2^2)^2\cdot (x^2)^2\cdot y^{\frac{3}{5}}} {(2^5)^{\frac{2}{5}}\cdot (x^5)^{\frac{2}{5}}\cdot (y^4)^{\frac{2}{5}}}

. . =\;\frac{2^4\cdot x^4\cdot y^{\frac{3}{5}}}{2^2\cdot x^2\cdot y^{\frac{8}{5}}}

. . =\;2^{4-2}\cdot x^{4-2}\cdot y^{\frac{3}{5}-\frac{8}{5}}

. . =\;2^2\cdot x^2\cdot y^{-1}

. . =\;\frac{4x^2}{y}

 

[sweer6]

what i quite don't understand about this question "[32x^5y^4)^2/5/(4x^2)^2y^3/5]^-1" is after the first raise to the power of two (bold above) is the index devided by 5 and so on or is it the question so far that is divided by 5 and so on.