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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`
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Solution
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.
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