Question

If $$a : b :: c : d$$, show that $$\frac{(a + c)^3}{(b + d)^3} = \frac{a(a - c)^2}{b(b - d)^2}$$.

[Hint : Use k-method.]

Answer 1

Let $$\frac{a}{b} = \frac{c}{d} = k$$, so $$a = bk$$ and $$c = dk$$.

$$\text{LHS} = \frac{(bk + dk)^3}{(b + d)^3}$$

$$= \frac{k^3(b + d)^3}{(b + d)^3}$$

$$= k^3$$

$$\text{RHS} = \frac{bk(bk - dk)^2}{b(b - d)^2}$$

$$= \frac{bk \cdot k^2(b - d)^2}{b(b - d)^2}$$

$$= k^3$$

$$\text{LHS} = \text{RHS}$$