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.

 A: $4n^{\frac{3}{2}}=8^{-\frac{1}{3}} \iff 4n^{\frac{3}{2}}=\frac{1}{2} \iff n^{\frac{3}{2}}=\frac{1}{8} \iff n^3=\frac{1}{64} \iff n= \frac{1}{4}$
A: $${ \left( 4n \right)  }^{ \frac { 3 }{ 2 }  }={ \left( 8 \right)  }^{ -\frac { 1 }{ 3 }  }\\ \left( { \left( 4n \right)  }^{ \frac { 3 }{ 2 }  } \right) ^{ 2/3 }=\left( { \left( { 2 }^{ 3 } \right)  }^{ -\frac { 1 }{ 3 }  } \right) ^{ 2/3 }\\ 4n=2^{ -\frac { 2 }{ 3 }  }\\ n=\frac { 2^{ -\frac { 2 }{ 3 }  } }{ 4 } =\frac { 1 }{ 4\sqrt [ 3 ]{ 4 }  } $$
A: We have that $4n^{3/2}=8^{-1/3}$. Proceed as follows, to find $n$.


*

*Simplify $8^{-1/3}=1/\sqrt[3]{8}=1/2$

*Divide both sides by 4 to get $n^{3/2}=1/8$. This is the next step, bearing in mind the order of operations, which is often remembered with the phrase BODMAS(/BIDMAS) (Brackets, Other(/Indices), Division, Multiplication, Addition, Subtraction). That is, if we were to add in unnecessary brackets to emphasize the order, we would write $4(n^{3/2})=1/2$, and we can see dividing by 4 is the next available choice to isolate $n$.

*Raise both sides to the power 2/3 (or if you like, take the "3/2"th root) of both sides, and simplify, to get $n=(1/8)^{2/3}=1/8^{2/3}=1/\sqrt[3]{8}^2=1/2^2=1/4$

A: 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
$n = \dfrac{1}{4}$.
A: We have $$(4n)^\frac{3}{2} = (8)^{-\frac{1}{3}}$$
So we write as $4^{3/2} n^{3/2} = (2^{3})^{-1/3}$
Now writing as
$(2^{2})^{3/2} n^{3/2}= 2^{-1}$
we get as
$8n^{3/2}=1/2$
so $n^{3/2}=1/16$
Now squaring both sides we get
$n^{3}= (1/16) (1/16)$
A: Your equation is: $\{4 n\}^\frac32 = \frac12$.(I think you understood this)
Now, write $4=2^2$ in the left side. Then the equation looks like
$(2^2)^\frac32 \times n^\frac32=\frac12$ 
$\Rightarrow $$n^\frac32 =$$\frac1{16}$
$\Rightarrow n=(\frac1{16})^\frac23$$=\frac14\times 2^{-\frac23}$
A: $$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$$
