If `h , c ,V` are respectively the height, the curved surface and the volume of a cone, prove that `3piV h^3-C^2h^2+9V^2=0.`

Explanation

Let `r` and `l` denote respectively the radius of the base and slant height of the cone.
`l=sqrt(r^2+h^2)`
`V=1/3pir^2h`
`C=pirl`
`therefore 3piVh^3−C^2h^2+9V^2`
`=3pi×1/3pir^2h×h^3-(pirl)^2h^2+9×(1/3pir^2h)^2`
`=pi^2r^2h^4−pi^2r^2l^2h^2+pi^2r^4h^2`
`=pi^2r^2h^4−pi^2r^2h^2(r^2+h^2)+pi^2r^4h^2`
`=pi^2r^2h^4−pi^2r^4h^2−pi^2r^2h^4+pi^2r^4h^2=0`
Hence, the answer is `0`.