If $$h, c, v,$$ are the height, the curved surface area and the volume of a cone, then $$3\pi vh^3$$ is

$$c\ =\ \pi\ rl\ =\ \pi\ r\sqrt{\ r^2+h^2}$$

$$v\ =\ \frac{1}{3}\pi\ r^2h$$

$$3\pi\ vh^3\ =\ 3\pi\ \left(\frac{1}{3}\pi\ r^2h\right)h^3\ =\ \pi\ ^2r^2h^4$$

A) $$9v^2-c^2h^3\ =\ 9\left(\frac{1}{3}\pi\ r^2h\right)^2-\pi\ ^2r^2\left(h^2+r^2\right)h^{3\ }=\ \pi\ ^2r^4h^2-\pi\ ^2r^2h^5-\pi\ ^2r^4h^3$$ 

B)$$c^2h^2-9v^2\ =\ \pi\ ^2r^2\left(h^2+r^2\right)h^2-9\left(\frac{1}{3}\pi\ r^2h\right)^2=\ \pi\ ^2r^2h^4+\pi\ ^2r^4h^{2\ }-\ \pi\ ^2r^4h^2\ =\ \pi\ ^2r^2h^4$$ = $$3\pi\ vh^3$$

option C and D is not equal to $$3\pi\ vh^3$$

Answer is option B.