Question

If x, y and z are in continued proportion, Prove that:

`x/(y^2.z^2) + y/(z^2.x^2) + z/(x^2.y^2) = 1/x^3 + 1/y^3 + 1/z^3`

Answer 1

Given: x, y and z are in continued proportion.

∴ `x/y = y/z`

⇒ y² = xz

To prove: `x/(y^2.z^2) + y/(z^2.x^2) + z/(x^2.y^2) = 1/x^3 + 1/y^3 + 1/z^3`

Proof: Solving L.H.S.:

`x/(y^2.z^2) + y/(z^2.x^2) + z/(x^2.y^2)`

⇒ `(x^3 + y^3 + z^3)/(x^2.y^2.z^2)`

⇒ `(x^3 + y^3 + z^3)/(x^2.xz.z^2)`

⇒ `(x^3 + y^3 + z^3)/(x^3.z^3)`

⇒ `x^3/(x^3.z^3) + y^3/(x^3.z^3) + z^3/(x^3.z^3)`

⇒ `1/z^3 + y^3/(xz)^3 + 1/x^3`

⇒ `1/z^3 + y^3/(y^2)^3 + 1/x^3`

⇒ `1/z^3 + y^3/y^6 + 1/x^3`

⇒ `1/z^3 + 1/y^3 + 1/x^3`

Since L.H.S. = R.H.S.

Hence proved.