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 so

`l= r^2+h^2`

`V=(1/3)pi(r^2)h`

`C=πrl`

Therefore,
`LHS=3piV(h^3)−(C^2)(h^2)+9V^2`

`=3pi{(1/3)pi(r^2)h}h^3-{(pirl)^2}h^2+9{(1/3)pi(r^2)h}^2`

`=(pi^2)(r^2)(h^4)-(pi^2)(r^2)(l^2)(h^2)+(pi^2)(r^4)(h^2)`

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