<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] So the real problem we are trying to complete is to solve $4n^{\frac{3}{2}} = 8^{-\frac{1}{3}}$.
The way to do this is to first divide both sides by $4$ and get:
$n^{\frac{3}{2}} = \dfrac{8^{-\frac{1}{3}}}{4}$
Now, since $8 = 2^{3}$ and $4 = 2^{2}$, we can rewrite this as:
$n^{\frac{3}{2}} = \dfrac{(2^{3})^{-\frac{1}{3}}}{2^{2}}$
and in the numerator, since we have something with a power raised to another power, we can multiply the two powers to get:
$n^{\frac{3}{2}} = \dfrac{2^{3(-\frac{1}{3})}}{2^{2}}$
which gives
$n^{\frac{3}{2}} = \dfrac{2^{-1}}{2^{2}}$
Now, raising something to a negative exponent means you move it through the fraction either up or down depending on where it started.  So since $2^{-1}$ is in the numerator, we move it down to the denominator and remove the negative in the exponent, to get:
$n^{\frac{3}{2}} = \dfrac{1}{2^{1}2^{2}}$
Since we are multiplying two things raised to exponents with the same base in the denominator (the base is $2$), we can just add the exponents to get:
$n^{\frac{3}{2}} = \dfrac{1}{2^{3}}$
Finally, we can raise both sides to the $\frac{2}{3}$ power to get:
$(n^{\frac{3}{2}})^{\frac{2}{3}} = \left (\dfrac{1}{2^{3}} \right )^{\frac{2}{3}}$.
When you raise something with an exponent to another exponent, you can just multiply the exponents together to get:
$n^{\frac{3}{2} \cdot \frac{2}{3}} = \left (\dfrac{1}{2^{3}} \right )^{\frac{2}{3}}$
or
$n = \left (\dfrac{1}{2^{3}} \right )^{\frac{2}{3}}$
or
$n = \dfrac{1^{\frac{2}{3}}}{(2^{3})^{\frac{2}{3}}}$
Now, $1$ raised to any power is just $1$, so the numerator is $1$.  With the denominator, we again multiply exponents to get:
$n = \dfrac{1}{2^{3 \cdot {\frac{2}{3}}}}$
or
$n = \dfrac{1}{2^{2}}$
or
$n = \dfrac{1}{4}$.
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