<s>[INST] Simplify $\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} * \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$ to $\frac{\sqrt{mnc}}{a^9cmn}$ I need to simplify $$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot  \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$$
The solution provided is:  $\dfrac{\sqrt{mnc}}{a^9cmn}$.
I'm finding this challenging. I was able to make some changes but I don't know if they are on the right step or not:
First, I am able to simplify the left fractions numerator and the right fractions denominator:
$\sqrt{mn^3}=\sqrt{mn^2n^1}=n\sqrt{mn}$
$\sqrt{m^2c^4}=m\sqrt{c^2c^2} = mcc$
So my new expression looks like:
$$\frac{n\sqrt{mn}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{mcc}$$
From this point I'm really at a loss to my next steps. If I multiply them both together I get:
$$\frac{(n\sqrt{mn})(a^{-7}n^{-2})}{(a^2\sqrt{c^{-3}})(mcc)}.$$
Next, I was thinking I could multiply out the radical in the denominator but I feel like I need to simplify what I have before going forwards.
Am I on the right track? How can I simplify my fraction above in baby steps? I'm particularly confused by the negative exponents.
How can I arrive at the solution $\dfrac{\sqrt{mnc}}{a^9cmn}$?
 [/INST] Recall that $a^{-x}=\frac{1}{a^x}$, $a^{1/x}=\sqrt[x]{a}$, and $a^na^m=a^{m+n}$
Always convert everything to exponents first, then use arithmetic
$$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}}\frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}=m^{1/2}\ n^{3/2}\ a^{-2}\ c^{3/2}\ a^{-7}\ n^{-2}\ m^{-2/2}\ c^{-4/2}$$
Rearrange
$$m^{1/2}\ n^{3/2}\ a^{-2}\ c^{3/2}\ a^{-7}\ n^{-2}\ m^{-2/2}\ c^{-4}=(m^{1/2}\ m^{-2/2})\ (n^{3/2}\ n^{-2})\ (a^{-2}\ a^{-7})\ (c^{3/2}\ c^{-4/2})$$
Simplify
$$(m^{1/2-1})\ (n^{3/2-2})\ (a^{-2-7})\ (c^{3/2-2})=(m^{-1/2})\ (n^{-1/2})(a^{-9})(c^{-1/2})=\frac{1}{a^9\sqrt{mnc}}$$
Multiply by $\frac{\sqrt{mnc}}{\sqrt{mnc}}$
$$\frac{1}{a^9\sqrt{mnc}}\frac{\sqrt{mnc}}{\sqrt{mnc}}=\frac{\sqrt{mnc}}{a^9mnc}$$
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