The expression
\[\sin^3 2x \cos 6x + \cos^3 2x \sin 6x\]can be written in the equivalent form $a \sin bx$ for some positive constants $a$ and $b.$  Find $a + b.$
Applying the triple angle formula, we get
\begin{align*}
\sin^3 2x \cos 6x + \cos^3 2x \sin 6x &= \left( \frac{3}{4} \sin 2x - \frac{1}{4} \sin 6x \right) \cos 6x + \left( \frac{3}{4} \cos 2x + \frac{1}{4} \cos 6x \right) \sin 6x \\
&= \frac{3}{4} \sin 2x \cos 6x + \frac{3}{4} \cos 2x \sin 6x.
\end{align*}Then by the angle addition formula,
\[\frac{3}{4} \sin 2x \cos 6x + \frac{3}{4} \cos 2x \sin 6x = \frac{3}{4} \sin (2x + 6x) = \frac{3}{4} \sin 8x.\]Thus, $a + b = \frac{3}{4} + 8 = \boxed{\frac{35}{4}}.$