If the height, curved surface area and the volume of a cone are h, c and v respectively, then 3 π v h³-c² h²… · ABC26GN4962

Volume of the cone, $v=\frac{1}{3} \pi r^{2} h$. Curved surface area, $c=\pi r l=\pi r \sqrt{r^{2}+h^{2}}$ $$\begin{aligned} & \Rightarrow c^{2}=\pi^{2} r^{2}\left(r^{2}+h^{2}\right) . \\ & \therefore 3 \pi v h^{3}-c^{2} h^{2}+9 v^{2} \\ & =3 \pi \times \frac{1}{3} \pi r^{2} h \times h^{3}-\pi^{2} r^{2}\left(r^{2}+h^{2}\right) h^{2}+9 \times \frac{1}{9} \pi^{2} r^{4} h^{2} \\ & \quad=\pi^{2} r^{2} h^{4}-\pi^{2} r^{4} h^{2}-\pi^{2} r^{2} h^{4}+\pi^{2} r^{4} h^{2}=0 . \end{aligned}$$