<s>[INST] 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.
 [/INST] 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
$$ \frac{S_c}{S_s} \ \ = \ \ \frac{6 \ V ^{2/3}}{(36 \ \pi)^{1/3} \ V ^{2/3}} \ \  = \ \ \frac{6   }{6^{2/3} \ \pi^{1/3} } \ \ = \ \ \left(\frac{6   }{  \pi  } \right)^{1/3} \ \ = \ \ \frac{6^{1/3}   }{  \pi^{1/3}  } \ \ \text{or} \ \ \frac{\sqrt[3]{6}   }{ \sqrt[3]{ \pi}  } \ \ .  $$
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