<s>[INST] 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.
 [/INST] You have:
$$\begin{align*}
\frac{(3x^{3/2}y^3)^{-2}}{(x^2y^{-1/2})} &= \frac{1}{(3x^{3/2}y^3)^2}\times \frac{1}{x^2} \times y^{1/2}\\
&= \frac{1}{(3^2)(x^{3/2})^2(y^3)^2}\times \frac{1}{x^2}\times\frac{y^{1/2}}{1}\\
&= \frac{1}{9x^3y^6}\times\frac{1}{x^2}\times \frac{y^{1/2}}{1}\\
&= \frac{1}{9x^3x^2}\times\frac{y^{1/2}}{y^6}\\
&= \frac{1}{9x^5}\times \frac{1}{y^{11/2}}\\
&= \frac{1}{9x^5y^{11/2}}.
\end{align*}$$
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