Notice that $(a-bi)^{3}=a^{3}-3a^{2}bi-3ab^{2}+b^{3}i=(a^{3}-3ab^{2})+(b^{3}-3ba^{2})i=36+i(28i)=8$ so that $a-bi=2+i$. Additionally $(a+bi)^{3}=a^{3}+3a^{2}bi-3ab^{2}-b^{3}i=(a^{3}-3ab^{2})-(b^{3}-3ba^{2})i=36-i(28i)=64$. It follows that $a-bi=2\omega$ and $a+bi=4\omega^{\prime}$ where $\omega, \omega^{\prime}$ are third roots of unity. So $a=\omega+2\omega^{\prime}$. From the triangle inequality $|a| \leq|\omega|+\left|2\omega^{\prime}\right|=3$, with equality when $\omega$ and $\omega^{\prime}$ point in the same direction (and thus $\omega=\omega^{\prime}$ ). It follows that $a=3,3\omega, 3\omega^{2}$, and so $a=3,-\frac{3}{2}+\frac{3i\sqrt{3}}{2},-\frac{3}{2}-\frac{3i\sqrt{3}}{2}$.