How to simplify $\frac{(3x^{3/2}y^3)} {(x^2y^{-1/2})}^{-2}$? I can't figure this one out on my own either
$$\frac{(3x^{3/2}y^3)} {(x^2y^{-1/2})}^{-2}$$
I am a little confused on all the small rules at play here but I know that a negative exponent will flip a fraction so I square the top and then flip it. But before that I should work in the parentheses first since that is the order of operataions. I am not sure how to cancel out each of the numbers though I am a little confused what a negative fraction in the numerator does to a larger positive exponent in the denominator.
 A: You have:
$$\begin{align*}
\frac{(3x^{3/2}y^3)^{-2}}{(x^2y^{-1/2})} &= \frac{1}{(3x^{3/2}y^3)^2}\times \frac{1}{x^2} \times y^{1/2}\\
&= \frac{1}{(3^2)(x^{3/2})^2(y^3)^2}\times \frac{1}{x^2}\times\frac{y^{1/2}}{1}\\
&= \frac{1}{9x^3y^6}\times\frac{1}{x^2}\times \frac{y^{1/2}}{1}\\
&= \frac{1}{9x^3x^2}\times\frac{y^{1/2}}{y^6}\\
&= \frac{1}{9x^5}\times \frac{1}{y^{11/2}}\\
&= \frac{1}{9x^5y^{11/2}}.
\end{align*}$$
A: $$\frac{(3x^{3/2}y^3)} {(x^2y^{-1/2})}^{-2}=\frac{3^{-2}\cdot x^{-3}\cdot y^{-6}}{x^2 \cdot y^{\frac{-1}{2}}}=\frac{1}{9}\cdot x^{(-3-2)}\cdot y^{\left(-6+\frac{1}{2}\right)}$$
A: I'd distribute the power upstairs first ( $(ab)^x=a^xb^x$)
$$
(3x^{3/2} y^3)^{-2}= {3^{-2} (x^{3/2})^{-2} (y^3)^{-2}}
$$
Now, on the right hand side of the above  use $(a^x)^y=a^{xy}$
$$
3^{-2} (x^{3/2})^{-2} (y^3)^{-2}=3^{-2} x^{(3/2)\cdot(-2)}y^{ 3\cdot(-2)} =3^{-2} x^{-3}y^{-6}
$$
So you have
$$
3^{-2} x^{-3}y^{-6}\over x^2 y^{-1/2}
$$
Now use ${a^x\over a^y}=a^{x-y}$ 
$$
{3^{-2} \color{darkgreen}{x^{-3}}\color{maroon}{y^{-6}}\over\color{darkgreen}{ x^2}\color{maroon} {y^{-1/2}} }=3^{- 2}\color{darkgreen}{x^{-3-2}}\color{maroon}{y^{-6-(-1/2)}}=3^{-1/2}x^{-5}y^{-11/2}. 
$$
Then "pretty it up"
$$
{1\over 9x^5y^{11/2}}.
$$
