Question

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$ prove that:
$\left(bdf\right).{(\frac{a+b}{a}+\frac{c+d}{d}+\frac{e+f}{f})}^3=27(a+b)(c+d)(e+f)$

Answer 1

Given:- $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$

To proof:- $\left(bdf\right)\cdot {(\frac{a+b}{b}+\frac{c+d}{d}+\frac{e+f}{f})}^3=27(a+b)(c+d)(e+f)$

Proof:-

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

$\Rightarrow a=bk,c=dk,e=fk$

Now,

$\left(bdf\right)\cdot {(\frac{a+b}{b}+\frac{c+d}{d}+\frac{e+f}{f})}^3=27(a+b)(c+d)(e+f)$

$\left(bdf\right)\cdot {((1+\frac{a}{b})+(1+\frac{c}{d})+(1+\frac{e}{f}))}^3=27(bk+b)(dk+d)(fk+f)$

$\left(bdf\right)\cdot {(1+k+1+k+1+k)}^3=27\left(bdf\right){(1+k)}^3$

$\left(bdf\right)\cdot {(3+3k)}^3=27\left(bdf\right)\cdot {(1+k)}^3$

$\left(bdf\right)\cdot 3^3{(1+k)}^3=27\left(bdf\right)\cdot {(1+k^3)}^3$

$27\left(bdf\right)\cdot {(1+k)}^3=27\left(bdf\right)\cdot {(1+k)}^3$

L.H.S. $=$ R.H.S.

Hence proved.