If a, b, c, d are in continued proportion, prove that :

(i) $\dfrac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} = \dfrac{a}{d}$

(ii) (a² - b²)(c² - d²) = (b² - c²)²

(iii) (a + d)(b + c) - (a + c)(b + d) = (b - c)²

(iv) a : d = triplicate ratio of (a - b) : (b - c)

(v) $\big(\dfrac{a - b}{c} + \dfrac{a - c}{b}\big)^2 - \big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\big)^2 = (a - d)^2\big(\dfrac{1}{c^2} - \dfrac{1}{b^2}\big).$

If a : b : : c : d, prove that

$\begin{matrix} \text{(i)} & \dfrac{2a + 5b}{2a - 5b} = \dfrac{2c + 5d}{2c - 5d}. \\[0.5em] \text{(ii)} & \dfrac{5a + 11b}{5c + 11d} = \dfrac{5a - 11b}{5c - 11d}. \\[0.5em] \text{(iii)} & (2a + 3b)(2c - 3d) \ & = (2a - 3b)(2c + 3d). \\ \text{(iv)} & (la + mb) : (lc + md) \ & : : (la - mb) : (lc - md). \end{matrix}$

$\dfrac{8a - 5b}{8c - 5d} = \dfrac{8a + 5b}{8c + 5d}, \text{ prove that } \dfrac{a}{b} = \dfrac{c}{d}.$

If (4a + 5b)(4c - 5d) = (4a - 5b)(4c + 5d), prove that a, b, c, d are in proportion.