This is the equation, please help me. I don't have any idea what to do with this. Thank you.
√((32)^((4)/(5))(2^(n+1))^(4))/8^3 (16)^(n-1)
That is a whole fraction within a radical sign all over √6:(9)
$$\frac{\sqrt{\frac{32^{\frac45}(2^{(n+1)})^4}{8^316^{(n-1)}}}}{\sqrt[6] 9}$$
First let's simplify the numerator:
$\sqrt{\dfrac{32^{4/5}\cdot(2^{n+1})^4}{8^3 \cdot 16^{n-1}}} =\sqrt{\dfrac{(2^5)^{4/5}\cdot(2^{n+1})^4}{(2^3)^3 \cdot (2^4)^{n-1}}} =\sqrt{\dfrac{2^4\cdot 2^{4n+4}}{2^9 \cdot 2^{4n-4}}} =\sqrt{2^{4-9+(4n+4)-(4n-4)}}=\sqrt{2^3}=2^{\frac32}$.
The denominator:
$\sqrt[6]9=9^{\frac16}=(3^2)^{\frac16}=3^{\frac13}$.
So, your expression is $\dfrac{2^\frac32}{3^\frac13}$. You can rewrite this as $\dfrac{2^\frac96}{3^\frac26}=\sqrt[6]{\dfrac{2^9}{3^2}}=\sqrt[6]{\dfrac{512}{9}}$.