Answer :
..
Given
[tex]\begin{gathered} \frac{dl}{dt}(rate,of,increase,in,side)=6cms^{-1} \\ l=9cm \end{gathered}[/tex]To Determine: The rate of increase in volume
The volume of a cube is given as
[tex]V=l^3[/tex]The rate of change of volume with respect to the length is the derivative of the volume function. So the derivative is as calculated below
[tex]\begin{gathered} \frac{dV}{dl}=3l^2 \\ =3\times(9cm)^2=3\times81cm^2=243cm^2 \end{gathered}[/tex]The rate of increase in volume would be
[tex]\frac{dV}{dt}(rate,of,increase,in,volume)=\frac{dV}{dl}\times\frac{dl}{dt}[/tex][tex]\begin{gathered} \frac{dV}{dt}=243cm^2\times6cms^{-1} \\ \frac{dV}{dt}=1458cm^3s^{-1} \end{gathered}[/tex]Hence, the rate of increase in volume is 1458cm³ per seconds