<s>[INST] Can someone show me HOW to do this, I don't just want the answer $4n$ to the power of $3$ over $2 = 8$ to the power of negative $1$ over $3$
Written Differently for Clarity:
$$(4n)^\frac{3}{2} = (8)^{-\frac{1}{3}}$$


EDIT
Actually, the problem should be solving $4n^{\frac{3}{2}} = 8^{-\frac{1}{3}}$.  Another user edited this question for clarity, but they edited it incorrectly to add parentheses around the right hand side, as can be seen above.

 [/INST] $$4n^{\frac32} = 8^{-\frac{1}{3}}= \frac{1}{8^{\frac{1}{3}}}$$
$$4\sqrt{n^3}= \frac{1}{\sqrt[3]{8}}= \frac{1}{2}$$
$$\sqrt{n^3}= \frac{1}{8}$$
$$\left(\sqrt{n^3}\right)^2= \left(\frac{1}{8}\right)^2$$
$$\left|n^3\right|= \frac{1}{8^2}=\frac{1}{64}$$
Note that $n^{\frac32}$ or $\sqrt{n^3}$ implies that $n^3\geq 0$ and we can drop the absolute value bars. So now we have
$$n^3=\frac{1}{64}$$
$$\sqrt[3]{n^3}=\sqrt[3]{\frac{1}{64}}$$
$$n=\frac{1}{\sqrt[3]{64}}=\frac14$$
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