If $\dfrac{x}{a} =\dfrac{y}{b} = \dfrac{z}{c},$ prove that

$\begin{array}{ll} \text{(i)} & \dfrac{x^3}{a^2} + \dfrac{y^3}{b^2} + \dfrac{z^3}{c^2} = \dfrac{(x + y + z)^3}{(a + b + c)^2} \\ \text{(ii)} & \Big(\dfrac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3z}\Big)^3 = \dfrac{xyz}{abc} \\ \text{(iii)} & \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} \ & + \dfrac{cz - ax}{(c + a)(z - x)} = 3. \end{array}$

If $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f}$, prove that

$\begin{array}{ll} \text{(i)} & (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) \ & = (ab + cd + ef)^2 \\ \text{(ii)} & \dfrac{(a^3 + c^3)^2}{(b^3 + d^3)^2} = \dfrac{e^6}{f^6} \\ \text{(iii)} & \dfrac{a^2}{b^2} + \dfrac{c^2}{d^2} + \dfrac{e^2}{f^2} \ & = \dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf} \\ \text{(iv)} & bdf\Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^3 \ & = 27(a + b)(c + d)(e + f) \end{array}$

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

$\begin{array}{ll} \text{(i)} & (5a + 7b)(2c - 3d) \ & = (5c + 7d)(2a - 3b) \\ \text{(ii)} & (ma + nb) : b \ & = (mc + nd) : d \\ \text{(iii)} & (a^4 + c^4) : (b^4 + d^4) \ & = a^2c^2 : b^2d^2 \\ \text{(iv)} & \dfrac{a^2 + ab}{c^2 + cd} = \dfrac{b^2 - 2ab}{d^2 - 2cd} \\[1em] \text{(v)} & \dfrac{(a + c)^3}{(b + d)^3} = \dfrac{a(a - c)^2}{b(b - d)^2} \\[1em] \text{(vi)} & \dfrac{a^2 + ab + b^2}{a^2 - ab + b^2} \ & = \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2} \\ \text{(vii)} & \dfrac{a^2 + b^2}{c^2 + d^2} = \dfrac{ab + ad - bc}{bc + cd - ad} \\[1em] \text{(viii)} & abcd\Big(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + \dfrac{1}{d^2} \Big) \ & = a^2 + b^2 + c^2 + d^2. \end{array}$

If x, y, z are in continued proportion, prove that : $\dfrac{(x + y)^2}{(y + z)^2} = \dfrac{x}{z}.$