Question

If $a, b, c$ are in continued proportion, prove that, $a^{2} b^{2} c^{2} (a^{- 4} + b^{- 4} + c^{- 4}) = b^{- 2} (a^{- 4} + b^{- 4} + c^{- 4}) .$

Accepted Answer

Given,

$a^{2} b^{2} c^{2} (a^{- 4} + b^{- 4} + c^{- 4}) = b^{- 2} (a^{4} + b^{4} + c^{4})$

$LHS = a^{2} b^{2} c^{2} (a^{- 4} b^{- 4} c^{- 4})$

We can write it as,

$= a^{2} b^{2} c^{2} [\frac{1}{a^{4}} + \frac{1}{b^{4}} + \frac{1}{c^{4}}]$

So we get

$= \frac{a^{2} b^{2} c^{2}}{a^{4}} + \frac{a^{2} b^{2} c^{2}}{b^{4}} + \frac{a^{2} b^{2} c^{2}}{c^{4}}$
By further calculation

$= \frac{b^{2} c^{2}}{a^{2}} + \frac{c^{2} a^{2}}{b^{2}} + \frac{a^{2} b^{2}}{c^{2}}$

It can be written as

$= \frac{(c k)^{2} c^{2}}{{(c k^{2})}^{2}} + \frac{c^{2} {(c k^{2})}^{2}}{(c k)^{2}} + \frac{{(c k^{2})}^{2} (c k)^{2}}{c^{2}}$

$= \frac{c^{2} k^{2} c^{2}}{c^{2} k^{4}} + \frac{c^{2} c^{2} k^{4}}{c^{2} k^{2}} + \frac{c^{2} k^{4} c^{2} k^{2}}{c^{2}}$

On further simplification

$= \frac{c^{2}}{k^{2}} + \frac{c^{2} k^{2}}{1} + \frac{c^{2} k^{6}}{1}$

Taking out the common terms

$= c^{2} [\frac{1}{k^{2}} + k^{2} + k^{6}]$

$= \frac{c^{2}}{k^{2}} [1 + k^{4} + k^{8}]$

$RHS = b^{- 2} [a^{4} + b^{4} + c^{4}]$

We can write it as
$= \frac{1}{b^{2}} [a^{4} + b^{4} + c^{4}]$

So we get

$= \frac{1}{(c k)^{2}} [{(c k^{2})}^{4} + (c k)^{4} + c^{4}]$

By further calculation

$= \frac{1}{c^{2} k^{2}} [c^{4} k^{8} + c^{4} k^{4} + c^{4}]$

Taking out the common terms

$= \frac{c^{4}}{c^{2} k^{2}} [k^{8} + k^{4} + 1]$

$= \frac{c^{2}}{k^{2}} [1 + k^{4} + k^{8}]$

Therefore, it is proved that LHS=RHS.

M L Aggarwal Solutions for Exercise 2: Exercise 7.2

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Exercise 2: Exercise 7.2

Questions from M L Aggarwal Solutions for Exercise 2: Exercise 7.2 with Hints & Solutions

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