Find the value of the constants If $\displaystyle
\frac{\left( \frac{2x^2}{3a} \right)^{n-1}}
         {\left( \frac{3x}{a} \right)^{n+1}} =
\left( \frac{x}{4} \right)^3$, determine the values of the constants $a$ and $n$
I could find the value of $a$, i.e, $\displaystyle \frac{\sqrt{x^6 \times 3^{2n}}}{2^n x^n 2^5 }$ and substituted the same to find the value of $n$, to no avail 
The right values for $a$ is $\pm(3^6 2^{-11/2})$, and $n$ is equal to $6$
 A: \begin{align*}
  \frac{\left( \frac{2x^2}{3a} \right)^{n-1}}
       {\left( \frac{3x}{a} \right)^{n+1}} &=
  \left( \frac{x}{4} \right)^3 \\
  \frac{2^{n-1}}{3^{2n}a^{2n}} x^{n-3} &=
  \frac{x^3}{2^6}
\end{align*}
Comparing $x$ terms,
$$n-3=3 \implies n=6$$
Comparing coefficient of $x^3$,
$$\frac{2^5 a^2}{3^{12}}=\frac{1}{2^6} \implies a^{2}=\frac{3^{12}}{2^{11}} \implies a=\pm \frac{3^6}{2^{11/2}}$$
A: Expand the LHS as
$$
\frac{{\left( {\frac{{2x^{\,2} }}
{{3a}}} \right)^{n - 1} }}
{{\left( {\frac{{3x}}
{a}} \right)^{n + 1} }} = \frac{{2^{n - 1} x^{2\left( {n - 1} \right)} }}
{{3^{n - 1} a^{n - 1} }}\;\frac{{a^{n + 1} }}
{{3^{n + 1} x^{n + 1} }} = \frac{{2^{n - 1} a^2 x^{n - 3} }}
{{3^{2n} }}\;
$$
The equating it to the RHS, you shall impose that the exponent
of $x$ be equal on both sides, as well as the multiplying coefficients.
Therefore: 
$$
\begin{gathered}
  \frac{{2^{n - 1} a^2 x^{n - 3} }}
{{3^{2n} }}\; = \frac{{x^3 }}
{{4^3 }}\quad  \Rightarrow \quad \left\{ \begin{gathered}
  x^{n - 3}  = x^3  \hfill \\
  \frac{{2^{n - 1} a^2 }}
{{3^{2n} }} = \frac{1}
{{4^3 }} \hfill \\ 
\end{gathered}  \right. \hfill \\
   \Rightarrow \quad \left\{ \begin{gathered}
  n = 6 \hfill \\
  \frac{{2^5 a^2 }}
{{3^{12} }} = \frac{1}
{{4^3 }} \hfill \\ 
\end{gathered}  \right.\quad  \Rightarrow \quad \left\{ \begin{gathered}
  n = 6 \hfill \\
  a^2  = \frac{{3^{12} }}
{{2^5 4^3 }} = \frac{{3^{12} }}
{{2^{11} }} \hfill \\ 
\end{gathered}  \right.\quad  \Rightarrow \quad \left\{ \begin{gathered}
  n = 6 \hfill \\
  a =  \pm \sqrt 2 \frac{{3^6 }}
{{2^6 }} \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} 
$$
