If $H$, $S$, and $V$ are the height, curved surface area, and volume of a cone respectively, then find the value of:
$$3\pi V{H}^3+9V^2-S^2{H}^2$$

Let the base radius of the cone be $r$ and slant height be $l=\sqrt{r^2+H^2}$.
The formulas for volume and curved surface area are:

$$V=\frac{1}{3}\pi r^{2}H⟹9V^2={\pi }^2{r}^4{H}^2$$

$$S=\pi rl⟹S^2={\pi }^2{r}^2{l}^2={\pi }^2{r}^2(r^2+H^2)$$

Substitute these expressions into the target formula:

$$\text{Expression}=3\pi \left(\frac{1}{3}\pi r^{2}H\right)H^3+{\pi }^2{r}^4{H}^2-\left[{\pi }^2{r}^2(r^2+H^2)\right]H^2$$

$$\text{Expression}={\pi }^2{r}^2{H}^4+{\pi }^2{r}^4{H}^2-{\pi }^2{r}^4{H}^2-{\pi }^2{r}^2{H}^4$$

All terms cancel each other out:

$$\text{Expression}=0$$