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.


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
$n = \dfrac{1}{4}$.
