If three small solid metallic spheres of radii 1 cm, 2 cm, and 3 cm respectively are completely melted to form a single solid sphere, then what is the surface area of the sphere so formed?

Let r be the radius of the solid sphere so formed.

It is known that the volume of solids remains the same in case of the conversion of solids from one shape into another.

∴Volume of three smaller spheres = Volume of the big sphere

$\frac{4}{3}\mathrm{\pi }{\left(1\right)}^3+\frac{4}{3}\mathrm{\pi }{\left(2\right)}^3+\frac{4}{3}\mathrm{\pi }{\left(3\right)}^3=\frac{4}{3}{\mathrm{\pi r}}^3$

$\Rightarrow 1+8+27={\mathrm{r}}^3$

$\Rightarrow \mathrm{r}={\left(36\right)}^{\frac{1}{3}}$

∴ The surface area of the sphere = 4πr²

$=4\mathrm{\pi }{\left[{\left(36\right)}^{\frac{1}{3}}\right]}^2$

$=4\mathrm{\pi }{\left(6^2\right)}^{\frac{2}{3}}$

$=4\mathrm{\pi }{\left(6\right)}^{\frac{4}{3}}$

$=24\mathrm{\pi }{\left(6\right)}^{\frac{1}{3}}$

Thus, the surface area of the sphere so formed is $24\mathrm{\pi }{\left(6\right)}^{\frac{1}{3}} {\mathrm{cm}}^2$.