[Math] Simplify $\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} * \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$ to $\frac{\sqrt{mnc}}{a^9cmn}$

algebra-precalculus

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 $\dfrac{\sqrt{mnc}}{a^9cmn}$?

Best Answer

\begin{align} \frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}} &= \frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{1}{a^7n^{2} mc^2}\\ & = \frac{\sqrt{mn^3}}{a^2} \cdot \frac{\sqrt{c^{3}}}{a^7n^{2}mc^2} \\ & = \frac{\sqrt{m}n\sqrt{n}c\sqrt{c}}{a^9 n^{2}m c^2} \\ & = \frac{\sqrt{nmc}}{a^9 n m c} \\ \end{align}

Related Question