Three metal cubes with edges 6cm, 8cm, 10cm respectively are melted together and formed into a single cube.

Let ‘a’ be the length of each edge of the new cube.

Then $a^{3}=\left(6^{3}+8^{3}+10^{3}\right) \mathrm{cm}^{3}$

$\Rightarrow a^{3}=1728$

⇒ a = 12

Therefore, Volume of the new cube $=a^{3}=1728 \mathrm{~cm}^{3}$

Surface area of the new cube $=6 \mathrm{a}^{2}=6 *(12)^{2}=864 \mathrm{~cm}^{2}$

Diagonal of the newly formed cube $=\sqrt{3} \mathrm{a}=12 \sqrt{3} \mathrm{~cm}$