A cube and a sphere have equal volume. What is the ratio of their surface areas? The answer is supposed to be $$ \sqrt[3]{6} : \sqrt[3]{\pi} $$
Since $$ \ a^3 = \frac{4}{3} \pi r^3  $$
I have expressed it as: $$ \ a = \sqrt[3]{ \frac{4}{3} \pi r^3} $$
and, 
$$  \ 6 \left( \sqrt[3]{ \frac{4}{3} \pi r^3 } \right) ^2 : 4 \pi r^2 $$
But I am not really sure how to arrive at the desired result. I have tried to simplify it, but apparently I am missing some step in the process and come to a result that is far from the correct one.
Could you please help? Thank you.

A sphere with equal volume to a cube of side $a$ must have radius $r$:
$$\frac{4\pi r^3}{3} = a^3$$
So
$$r = \sqrt[3]{\frac{3}{4 \pi}}a$$
Now just take
$$\frac{4\pi r^2}{6a^2} = \frac{4\pi\sqrt[3]{\frac{9}{16 \pi^2}}a^2}{6a^2} = \frac{2}{3}\pi\sqrt[3]{\frac{9}{16 \pi^2}} = \sqrt[3]{\frac{9\cdot 8 \pi^3}{27 \cdot 16 \pi^2}} = \sqrt[3]{\pi/6}.$$
