Question

IF a/b = c/d = e/f , prove the above sum

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Answer 1

Given;  

$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$    then    $bdf[\frac{a+b}{b} +\frac{c+d}{d} + \frac{e+f}{f} ]^{3}=?$

Let,

$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k$

Then, $a=bk$ , $c=dk$ and $e=fk$

$\therefore bdf[\frac{a+b}{b} +\frac{c+d}{d} + \frac{e+f}{f} ]^{3}$

Plug all value in above equation;

$=bdf[\frac{bk+b}{b} +\frac{dk+d}{d} + \frac{fk+f}{f} ]^{3}$

$=bdf[\frac{b(k+1)}{b} +\frac{d(k+1)}{d} + \frac{f(k+1)}{f} ]^{3}$

$=bdf[(k+1)+(k+1)+(k+1)]^{3}$

$=bdf[3(k+1)]^{3}$

$=27bdf[k+1]^{3}$

$=27bdf[\frac{a}{b} +1]^{3}$    (∵k=$\frac{a}{b}$ )

$=27(a+b)(c+d)(e+f)$

$\boxed{\bold{bdf[\frac{a+b}{b} +\frac{c+d}{d} + \frac{e+f}{f} ]^{3}=27(a+b)(c+d)(e+f)}}$

Hence Proved !!!!

Answer 2

Answer:

Use this picture for solution