Question
If b is the mean proportional between a and c, prove that $\frac{a^2-b^2+c^2}{a^{-2}-b^{-2}+c^{-2}}$ = $b^4$.
✓
Answer
Since, b is the mean proportional between a and c. So, $b^2$ = ac.
L.H.S. $=\frac{a^2-b^2+c^2}{a^{-2}-b^{-2}+c^{-2}}$
$=\frac{a^2-b^2+c^2}{\frac{1}{a^2}-\frac{1}{b^2}+\frac{1}{c^2}} $
$ =\frac{\left(a^2-b^2+c^2\right)}{\frac{b^2 c^2-a^2 c^2+a^2 b^2}{a^2 b^2 c^2}} $
$ =\frac{a^2 b^2 c^2\left(a^2-b^2+c^2\right)}{b^2 c^2-b^4+a^2 b^2}$
$=\frac{b^4 \times b^2\left(a^2-b^2+c^2\right)}{b^2\left(c^2-b^2+a^2\right)} $
$b^4=\text { R.H.S. }$
Hence proved.
L.H.S. $=\frac{a^2-b^2+c^2}{a^{-2}-b^{-2}+c^{-2}}$
$=\frac{a^2-b^2+c^2}{\frac{1}{a^2}-\frac{1}{b^2}+\frac{1}{c^2}} $
$ =\frac{\left(a^2-b^2+c^2\right)}{\frac{b^2 c^2-a^2 c^2+a^2 b^2}{a^2 b^2 c^2}} $
$ =\frac{a^2 b^2 c^2\left(a^2-b^2+c^2\right)}{b^2 c^2-b^4+a^2 b^2}$
$=\frac{b^4 \times b^2\left(a^2-b^2+c^2\right)}{b^2\left(c^2-b^2+a^2\right)} $
$b^4=\text { R.H.S. }$
Hence proved.
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