Spherplex

I will assign $R_1, A_1, V_1$ for the radius, area and volume of the original sphere and $R_2, A_2, V_2$ for the radius, area and volume of the smaller spheres. That being said, we shall start with c, the ratio of the volume: $V_{ 1 }=4/3\pi r^{ 3 }\\ V_2=V_1/n$

Which give us the value of c. $c=\frac{V_1}{V_2}=\frac {V_1}{V_1/n}=n$

With the value of the volume in hands, the radius should be calculated as follows: $R_{ 1 }=r\\ R_{ 2 }=\sqrt [ 3 ]{ \frac { V_{ 2 } }{ \frac { 4\pi }{ 3 } } } =\sqrt [ 3 ]{ \frac { V_{ 1 } }{ \frac { 4n\pi }{ 3 } } } =\sqrt [ 3 ]{ \frac { \frac { 4\pi }{ 3 } r^{ 3 } }{ \frac { 4n\pi }{ 3 } } } =\frac { r }{ \sqrt [ 3 ]{ n } }$

Therefore: $b=\frac { R_{ 1 } }{ R_{ 2 } }= \frac { r }{ \frac { r }{ \sqrt [ 3 ]{ n } } } =\sqrt [ 3 ]{ n }$

The areas are given by these formulas: $A_{ 1 }=4\pi r^{ 2 }\\ A_{ 2 }=4\pi r_{ 2 }^{ 2 }=4\pi { \left( \frac { r }{ \sqrt [ 3 ]{ n } } \right) }^{ 2 }$

So a is calculated as follows: $a=\frac{nA_2 }{A_1}=\frac { 4n\pi { \left( \frac { r }{ \sqrt [ 3 ]{ n } } \right) }^{ 2 } }{ 4\pi r^{ 2 } } =\frac { n }{ { \sqrt [ 3 ]{ n } }^{ 2 } } =\quad \sqrt [ 3 ]{ n }$

$a+b+c=\quad \sqrt [ 3 ]{ n } +\sqrt [ 3 ]{ n } +n=2\sqrt [ 3 ]{ n } +n$

So n must be a perfect cube. 1 can't be an answer because the original sphere is divided into smaller spheres. The next perfect cube is 8, therefore: $2\sqrt [ 3 ]{ n } +n=2\sqrt [ 3 ]{ 8 } +8= 2\times 2+8=4+8=12$

$Let ~x~be~the~radius~of~the~smaller~spheres..\\Volume ~is~same.~~~\implies~n\dfrac 4 3 \pi*x^3=\dfrac 4 3 \pi*r^3. ~~\\\Large \therefore~~\dfrac r x =\sqrt[3]n....(1)\\a=\dfrac {n4\pi*x^2}{4\pi*r^2} =\dfrac {nx^2}{r^2}=\dfrac n {\sqrt[3](n^2)}=\sqrt[3]n\\b=\dfrac r x =\sqrt[3]n.\\c=\dfrac{\frac 4 3 \pi*r^3}{\frac 4 3 \pi*x^3}=n \\\text{If n=1, means we have no small spheres. Since we have n>1, }\\ \text{the smallest }n = 2^3=8~ for~ a,~b,~ to~ be~ integer.\\\therefore~a+b+c= \sqrt[3]8+\sqrt[3]8+8=~~~\Large \color{#D61F06}{12}$