2100.35 – Fiddling with Exponents

$$\begin{align*}a &= 2^{(x+2)} \\a^3 &= 2^{(x+2) \times 3} \\&=2^{(3x+6)}\end{align*}$$

So,

$$\begin{align*}\dfrac{a^3}{2^6} &= \dfrac{2^{(3x+6)}}{2^6} \\[1em]&= 2^{(3x+6-6)} \\[0.5em]&= 2^{3x} \\[0.5em]&= (2^3)^x \\[0.5em]&= 8^x\end{align*}$$

Therefore $8^x = \dfrac{a^3}{64}$.