Question

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

[Hint : Use k-method in each.]

Answer 1

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

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

Proof:

1. Let $$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$$, so $$a = bk$$, $$c = dk$$, $$e = fk$$.
2. $$\frac{a + b}{b} = \frac{bk + b}{b} = k + 1$$, $$\frac{c + d}{d} = \frac{dk + d}{d} = k + 1$$, $$\frac{e + f}{f} = \frac{fk + f}{f} = k + 1$$
3. $$\text{L.H.S.} = (bdf) \cdot ((k + 1) + (k + 1) + (k + 1))^3 = (bdf) \cdot [3(k + 1)]^3 = 27 bdf (k + 1)^3$$
4. $$\text{R.H.S.} = 27(bk + b)(dk + d)(fk + f) = 27 b(k + 1) d(k + 1) f(k + 1) = 27 bdf (k + 1)^3$$
5. $$\text{L.H.S.} = \text{R.H.S.}$$

Hence proved.