Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm, respectively are melted and formed into a single cube.

We have,

Edges of the cubes: $a_{1}=6 \mathrm{~cm}, a_{2}=8 \mathrm{~cm}$ and $a_{3}=10 \mathrm{~cm}$

Let the edge of the new cube so formed be $a$.

As,

Volume of the new cube so formed $=a_{1}^{3}+a_{2}^{3}+a_{3}{ }^{3}$

$\Rightarrow a^{3}=6^{3}+8^{3}+10^{3}$

$\Rightarrow a^{3}=216+512+1000$

$\Rightarrow a^{3}=1728$

$\Rightarrow a=\sqrt[3]{1728}$

$\therefore a=12 \mathrm{~cm}$

So, the edge of the new cube so formed is 12 cm.