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

Given,

x, y and z are in continued proportion.

$\therefore \frac{x}{y} = \frac{y}{z}$

⇒ y² = xz

To prove :

$\frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$

Solving L.H.S,

$$\begin{aligned}\Rightarrow \frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^2.y^2.z^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^2.xz.z^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^3.z^3} \\[1em] \Rightarrow \frac{x^3}{x^3.z^3} + \frac{y^3}{x^3.z^3} + \frac{z^3}{x^3.z^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{(xz)^3} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{(y^2)^3} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{y^6} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{1}{y^3} + \frac{1}{x^3}.\end{aligned}$$

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

Hence, proved that $\frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$.