<s>[INST] 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 $6$
 [/INST] \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 $x^3$,
$$\frac{2^5 a^2}{3^{12}}=\frac{1}{2^6} \implies a^{2}=\frac{3^{12}}{2^{11}} \implies a=\pm \frac{3^6}{2^{11/2}}$$
</s>