An edge of a variable cube is increasing at the rate of 3 cm/s.

Let x be the length of a side and V be the volume of the cube. Then,

$V=x^{3}$

$\therefore \frac{d V}{d t}=3 x^{2} \cdot \frac{d x}{d t}$ (By chain rule)

It is given that,

$\frac{d x}{d t}=3 \mathrm{~cm} / \mathrm{s}$

$\therefore \frac{d V}{d t}=3 x^{2}(3)=9 x^{2}$

Thus, when x = 10 cm,

$\frac{d V}{d t}=9(10)^{2}=900 \mathrm{~cm}^{3} / \mathrm{s}$

Hence, the volume of the cube is increasing at the rate of 900 cm3/s when the edge is 10 cm long.