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.

Unit volume assumed:
$$
1 = \frac{4}{3}\pi r^3 = a^3
$$
So we have
$$
r = \sqrt[3]{\frac{3}{4\pi}}  \\
\quad a = 1
$$
So the surface ratio is
$$
A_s = 4 \pi r^2 
= 4 \pi \left( \frac{3}{4\pi} \right)^{2/3}
= \sqrt[3]{4\pi \cdot 9} 
=\sqrt[3]{\pi} \, 6^{2/3} \\
A_c = 6
$$
So we get
$$
A_s : A_c = 
\sqrt[3]{\pi} : \sqrt[3]{6}
$$
