Question

If y is the mean proportional between x and z, prove that : $\frac{x^2-y^2+z^2}{x^{-2}-y^{-2}+z^{-2}}=y^4$.

Answer 1

Dear student,

$$\begin{gathered} Given:yismeanproportionalbetweenxandz \\ \Rightarrow y^2=xz--\left[1\right] \\ TakingL.H.Spart \\ \frac{x^2-y^2+z^2}{x^{-2}-y^{-2}+z^{-2}} \\ =\frac{x^2-y^2+z^2}{{\displaystyle \frac{1}{x^2}}-\frac{1}{y^2}+{\displaystyle \frac{1}{z^2}}}--[u\sin g{a}^{-m}=\frac{1}{a^m}] \\ =\frac{x^2-xz+z^2}{{\displaystyle \frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}}}--[u\sin g1] \\ =\frac{x^2-xz+z^2}{{\displaystyle \frac{z^2-xz+x^2}{x^2{z}^2}}} \\ =\frac{\cancel{\left(x^2-xz+z^2\right)}{x}^2{z}^2}{{\displaystyle \cancel{\left(z^2-xz+x^2\right)}}} \\ =x^2{z}^2 \\ ={\left(xz\right)}^2 \\ ={\left(y^2\right)}^2--[u\sin g1] \\ =y^4 \\ =R.H.S \end{gathered}$$

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