1. ## Serious algebra problems...

I am having a difficult problem solving a few algebra problems. I am not new to forums, but am new to this forum. Here is what I need help with.

The first part is all raised to the 3/2.

1.((16a^(-4)) ^3/2 * (4ab^(1/2))^-2
( b^8 )

I got an answer with a denominator. 4
a^8 b^13

According to the rules, is this correct?

2. 4y^(-2) -2x^(-1)
(xy)^(-1)

I am completely lost on this one. Not sure what to do after shifting the variables, to make them positive.

All help will be appreciated.

2. Originally Posted by russainred95

1.((16a^(-4)) ^3/2 * (4ab^(1/2))^-2
( b^8 )

I got an answer with a denominator. 4
a^8 b^13

According to the rules, is this correct?

2. 4y^(-2) -2x^(-1)
(xy)^(-1)

All help will be appreciated.
$\displaystyle 1)\;\; \frac{(16a^{-4})^{\frac{3}{2}}(4ab^{\frac{1}{2}})^{-2}}{b^8}$

$\displaystyle \frac{16^\frac{3}{2}}{b^8 \cdot (a^4)^{\frac{3}{2}} \cdot 4^2 \cdot a^2 \cdot b}$

$\displaystyle \frac{4^3}{b^9 \cdot a^8\cdot 4^2} = \frac{4}{b^9 \cdot a^8}$

$\displaystyle 2)\;\; \frac{4y^{-2} - 2 x^{-1}}{(xy)^{-1}}$

$\displaystyle \frac{4 y^{-2}}{(xy)^{-1}} - \frac{2x^{-1}}{(xy)^{-1}}$

$\displaystyle \frac{4 (xy)}{y^2} - \frac{2(xy)}{x}$

$\displaystyle \frac{4x}{y} - 2y$

$\displaystyle \frac{4x-2y^2}{y}$

3. Thank you very much. Are you using some sort of tool to solve those problems?