, 2017

Solution:
Notice that
$$\begin{aligned} (a-b i)^{3} & =a^{3}-3 a^{2} b i-3 a b^{2}+b^{3} i \\ & =\left(a^{3}-3 a b^{2}\right)+\left(b^{3}-3 b a^{2}\right) i \\ & =36+i(28 i) \\ & =8 \end{aligned}$$
so that $a-b i=2+i$. Additionally
$$\begin{aligned} (a+b i)^{3} & =a^{3}+3 a^{2} b i-3 a b^{2}-b^{3} i \\ & =\left(a^{3}-3 a b^{2}\right)-\left(b^{3}-3 b a^{2}\right) i \\ & =36-i(28 i) \\ & =64 \end{aligned}$$
It follows that $a-b i=2 \omega$ and $a+b i=4 \omega^{\prime}$ 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
$$a=3,-\frac{3}{2}+\frac{3 i \sqrt{3}}{2},-\frac{3}{2}-\frac{3 i \sqrt{3}}{2}$$