If the motor pumps water (density $= p$ ) continuously through a pipe of area of cross-section $A$ with velocity $v$, then mass flowing out per second
$m =Avp$ ...(i)
Rate of increase of kinetic energy
$=\frac{1}{2} m v^{2}=\frac{1}{2}(A v p) v^{2}$ ...(ii)
Mass $m$, flowing out per sec, can be increased to $m'$ by increasing $v$ to $v'$ then power increases from $P$ to $P'$.
$\frac{p'}{p}=\frac{\frac{1}{2} A \rho v^{' 3}}{\frac{1}{2} A \rho v^{3}}$
or $\frac{p'}{p}=\left(\frac{v'}{v}\right)^{3}$
Now, $\frac{m'}{m}=\frac{A \rho v'}{A \rho v}=\frac{v'}{v}$
As $m'=n m, v'=n v$
$\therefore \frac{p'}{p}=n^{3}$
$\Rightarrow p'=n^{3} P$