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.

Another way to approach this is to write the surface areas of each solid in terms of its volume:
cube -- $$ S_c \ = \ 6 \ a^2 \ \ , \ \ V_c \ = \ a^3 \ \ \Rightarrow \ \ a \ = \ V_c^{1/3} \ \ \Rightarrow \ \ S_c \ = \ 6 \ (V^{1/3})^2 \ = \ 6 \ V_c^{2/3} \ \ ; $$
sphere -- $$ S_s \ = \ 4 \ \pi  \ r^2 \ \ , \ \ V_s \ = \ \frac{4}{3}  \ \pi \ r^3 \ \ \Rightarrow \ \ r \ = \ \left(\frac{3 \ V_s}{4 \ \pi} \right)^{1/3} $$ $$ \Rightarrow \ \ S_s \ = \ 4 \ \pi \ \left[\left(\frac{3 \ V_s}{4 \ \pi} \right)^{1/3} \right]^2 \ = \ (4 \ \pi)^{1/3} \cdot 3^{2/3} \cdot V_s^{2/3} \ = \ \ (36 \ \pi)^{1/3} \ V_s^{2/3} \ \ . $$
The ratio of the surface areas is then
$$ \frac{S_c}{S_s} \ \ = \ \ \frac{6 \ V_c^{2/3}}{(36 \ \pi)^{1/3} \ V_s^{2/3}} \ \  ; $$
upon equating the volumes, we have
$$ \frac{S_c}{S_s} \ \ = \ \ \frac{6 \ V ^{2/3}}{(36 \ \pi)^{1/3} \ V ^{2/3}} \ \  = \ \ \frac{6   }{6^{2/3} \ \pi^{1/3} } \ \ = \ \ \left(\frac{6   }{  \pi  } \right)^{1/3} \ \ = \ \ \frac{6^{1/3}   }{  \pi^{1/3}  } \ \ \text{or} \ \ \frac{\sqrt[3]{6}   }{ \sqrt[3]{ \pi}  } \ \ .  $$
