If y is the mean proportional between x and z, prove that :

$\dfrac{x^2 - y^2 + z^2}{x^{-2} - y^{-2} + z^{-2}} = y^4$

Given, y is the mean proportional between x and z.

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

y² = xz

$\text{L.H.S. } = \dfrac{x^2 - y^2 + z^2}{x^{-2} - y^{-2} + z^{-2}} \\[1em] = \dfrac{x^2 - y^2 + z^2}{\dfrac{1}{x^2} - \dfrac{1}{y^2} + \dfrac{1}{z^2}} \\[1em] = \dfrac{x^2 - xz + z^2}{\dfrac{1}{x^2} - \dfrac{1}{xz} + \dfrac{1}{z^2}} \text{ (Substituting } y^2 = xz) \\[1em] = \dfrac{x^2 - xz + z^2}{\dfrac{z^2 - xz + x^2}{x^2z^2}} \\[1em] = x^2z^2\dfrac{x^2 - xz + z^2}{x^2 - xz + z^2} \\[1em] = x^2z^2 \\[1em] = (y^2)^2 \\[1em] = y^4 = \text{ R.H.S.}$

Hence, proved that $\dfrac{x^2 - y^2 + z^2}{x^{-2} - y^{-2} + z^{-2}} = y^4$.