# How to prove ?

## If $\frac{a}{b} = \frac{c}{d} = \frac{e}{f}$ prove that: $\frac{1}{27} {\left(\frac{a + b}{b} + \frac{c + d}{d} + \frac{e + f}{f}\right)}^{3} = \frac{\left(a + b\right) \left(c + d\right) \left(e + f\right)}{b \mathrm{df}}$

Apr 12, 2018

We start by expanding the LHS of the equation.

$\frac{1}{27} {\left(\frac{a}{b} + \frac{c}{d} + \frac{e}{f} + 3\right)}^{3}$

Let $x = \frac{a}{b} = \frac{c}{d} = \frac{e}{f}$. Then our LHS is $\frac{1}{27} {\left(3 x + 3\right)}^{3}$.

This can be further simplified: $\frac{1}{27} {\left(3 x + 3\right)}^{3} = \frac{1}{27} {\left(3 \cdot \left(x + 1\right)\right)}^{3} = \frac{27}{27} {\left(x + 1\right)}^{3} = {\left(x + 1\right)}^{3}$

We move to the RHS. Notice that

$\frac{\left(a + b\right) \left(c + d\right) \left(e + f\right)}{b \mathrm{df}} = \frac{a + b}{b} \cdot \frac{c + d}{d} \cdot \frac{e + f}{f}$
$= \left(\frac{a}{b} + 1\right) \left(\frac{c}{d} + 1\right) \left(\frac{e}{f} + 1\right) = {\left(x + 1\right)}^{3}$.

Indeed, both the LHS and RHS simplify to the same expression, so they must be equivalent.