Question
If $a, b, c$ are in continued proportion, prove that $: a^2 b^2 c^2 (a^{-4} + b^{-4} + c^{-4}) = b^{-2}(a^4 + b^4 + c^4)$
✓
Answer
Given : $a, b, c$ are in continued proportion.
$\frac{a}{b}=\frac{b}{c}= k$
$ \frac{a}{b}= k \therefore a = bk$
$ \frac{b}{c}= k \therefore b = ck$
$ \text { L.H.S. }= a ^2 b ^2 c ^2\left( a ^{-4}+ b ^{-4}+ c ^{-4}\right)$
$ \text { L.H.S. }=a^2 b^2 c^2\left[\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\right]$
$ \text { L.H.S. }=\frac{a^2 b^2 c^2}{a^4}+\frac{a^2 b^2 c^2}{b^4}+\frac{a^2 b^2 c^2}{c^4}$
$ \text { L.H.S. }=\frac{b^2 c^2}{a^2}+\frac{c^2 a^2}{b^2}+\frac{a^2 b^2}{c^2}$
$ \text { L.H.S. }=\frac{(c k)^2 \cdot c^2}{\left(c k^2\right)^2}+\frac{c^2\left(c k^2\right)^2}{(c k)^2}+\frac{\left(c k^2\right)^2(c k)^2}{c^2}$
$ \text { L.H.S. }=\frac{c^2 k^2 \cdot c^2}{c^2 k^4}+\frac{c^2 \cdot c^2 k^4}{c^2 k^2}+\frac{c^2 k^4 \cdot c^2 k^2}{c^2}$
$ \text { L.H.S. }=\frac{c^2}{k^2}+\frac{c^2 k^2}{1}+\frac{c^2 k^6}{1}$
$ \text { L.H.S. }=c^2\left[\frac{1}{k^2}+k^2+k^6\right]$
$ \text { L.H.S. }=\frac{c^2}{k^2}\left[1+k^4+k^8\right]$
$ \text { R.H.S. }=b^{-2}\left[a^4+b^4+c^4\right]$
$ \text { R.H.S. }=\frac{1}{b^2}\left[a^4+b^4+c^4\right]$
$ \text { R.H.S. }=\frac{1}{c^2 k^2}\left[c^4 k^8+c^4 k^4+c^4\right]$
$\text { R.H.S. }=\frac{c^4}{c^2 k^2}\left[k^8+k^4+1\right]$
$ \text { R.H.S. }=\frac{c^2}{k^2}\left[1+k^4+k^8\right]$
$ \therefore \text { L.H.S. }=\text { R.H.S. }$
Hence proved.
$\frac{a}{b}=\frac{b}{c}= k$
$ \frac{a}{b}= k \therefore a = bk$
$ \frac{b}{c}= k \therefore b = ck$
$ \text { L.H.S. }= a ^2 b ^2 c ^2\left( a ^{-4}+ b ^{-4}+ c ^{-4}\right)$
$ \text { L.H.S. }=a^2 b^2 c^2\left[\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\right]$
$ \text { L.H.S. }=\frac{a^2 b^2 c^2}{a^4}+\frac{a^2 b^2 c^2}{b^4}+\frac{a^2 b^2 c^2}{c^4}$
$ \text { L.H.S. }=\frac{b^2 c^2}{a^2}+\frac{c^2 a^2}{b^2}+\frac{a^2 b^2}{c^2}$
$ \text { L.H.S. }=\frac{(c k)^2 \cdot c^2}{\left(c k^2\right)^2}+\frac{c^2\left(c k^2\right)^2}{(c k)^2}+\frac{\left(c k^2\right)^2(c k)^2}{c^2}$
$ \text { L.H.S. }=\frac{c^2 k^2 \cdot c^2}{c^2 k^4}+\frac{c^2 \cdot c^2 k^4}{c^2 k^2}+\frac{c^2 k^4 \cdot c^2 k^2}{c^2}$
$ \text { L.H.S. }=\frac{c^2}{k^2}+\frac{c^2 k^2}{1}+\frac{c^2 k^6}{1}$
$ \text { L.H.S. }=c^2\left[\frac{1}{k^2}+k^2+k^6\right]$
$ \text { L.H.S. }=\frac{c^2}{k^2}\left[1+k^4+k^8\right]$
$ \text { R.H.S. }=b^{-2}\left[a^4+b^4+c^4\right]$
$ \text { R.H.S. }=\frac{1}{b^2}\left[a^4+b^4+c^4\right]$
$ \text { R.H.S. }=\frac{1}{c^2 k^2}\left[c^4 k^8+c^4 k^4+c^4\right]$
$\text { R.H.S. }=\frac{c^4}{c^2 k^2}\left[k^8+k^4+1\right]$
$ \text { R.H.S. }=\frac{c^2}{k^2}\left[1+k^4+k^8\right]$
$ \therefore \text { L.H.S. }=\text { R.H.S. }$
Hence proved.
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