The simplest way to do this is via a complex detour. (I'll do it for $a = 1$, it's straight-forward to adapt the solution to other radii.) Note that if $z = re^{i\theta}$, then $\newcommand{\imag}{\operatorname{Im}}\imag{z} = r\sin \theta$. The real and imaginary parts of a holomorphic function is harmonic, and since
$\sin^3 \theta = \frac34\sin\theta - \frac14 \sin 3\theta$
it makes sense to look at $f(z) = \frac34 z - \frac14 z^3$
If $|z| = 1$, then $\imag(f(z)) = \frac34\sin\theta - \frac14\sin 3\theta$, so the solution you want is
\begin{align} u = \imag(f(x+iy)) &= \imag\Big( \frac34(x+iy) - \frac14(x+iy)^3 \Big) \\ &= \frac34 y - \frac34 x^2 y + \frac14 y^3 \end{align}