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

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

(i) Let $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = k$

∴ x = ak, y = bk, z = ck.

$\text{L.H.S.} = \dfrac{x^3}{a^2} + \dfrac{y^3}{b^2} + \dfrac{z^3}{c^2} \\[0.5em] = \dfrac{a^3k^3}{a^2} + \dfrac{b^3k^3}{b^2} + \dfrac{c^3k^3}{c^2} \\[0.5em] = ak^3 + bk^3 + ck^3 \\[0.5em] =k^3(a + b + c).$

$\text{R.H.S.} = \dfrac{(x + y + z)^3}{(a + b + c)^2} \\[0.5em] = \dfrac{(ak + bk + ck)^3}{(a + b + c)^2} \\[0.5em] = \dfrac{k^3(a + b + c)^3}{(a + b + c)^2} \\[0.5em] = k^3(a + b + c).$

Since, L.H.S. = R.H.S.,

Hence proved, that

$\dfrac{x^3}{a^2} + \dfrac{y^3}{b^2} + \dfrac{z^3}{c^2} = \dfrac{(x + y + z)^3}{(a + b + c)^2}$

(ii) Let $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = k$

∴ x = ak, y = bk, z = ck.

$\text{L.H.S.} = \Big(\dfrac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3}\Big)^3 \\[1em] = \Big(\dfrac{a^2(ak)x + b^2(bk)y + c^2(ck)z}{a^3x + b^3y + c^3z}\Big)^3 \\[1em] = \Big(\dfrac{a^3xk + b^3yk + c^3zk}{a^3x + b^3y + c^3z}\Big)^3 \\[1em] = \dfrac{k^3(a^3x + b^3y + c^3z)^3}{(a^3x + b^3y + c^3z)^3} \\[1em] = k^3 \\[1em] = k \times k \times k \\[1em] = \dfrac{x}{a} \times \dfrac{y}{b} \times \dfrac{z}{c} \\[1em] = \dfrac{xyz}{abc} = \text{R.H.S.}$

Since, L.H.S. = R.H.S.

Hence proved, that $\Big(\dfrac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3z}\Big)^3 = \dfrac{xyz}{abc}$.

(iii) Let $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = k$

∴ x = ak, y = bk, z = ck.

$\text{L.H.S.} = \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} \\[1em] = \dfrac{a(ak) - b(bk)}{(a + b)((ak) - (bk))} + \dfrac{b(bk) - c(ck)}{(b + c)(bk - ck)} + \dfrac{c(ck) - a(ak)}{(c + a)(ck - ak)} \\[1em] = \dfrac{a^2k - b^2k}{k(a + b)(a - b)} + \dfrac{b^2k - c^2k}{k(b + c)(b - c)} + \dfrac{c^2k - a^2k}{k(c + a)(c - a)} \\[1em] = \dfrac{k(a^2 - b^2)}{k(a^2 - b^2)} + \dfrac{k(b^2 - c^2)}{k(b^2 - c^2)} + \dfrac{k(c^2 - a^2)}{k(c^2 - a^2)} \\[1em] = 1 + 1 + 1 \\[1em] = 3 = \text{R.H.S.}$

Since, L.H.S. = R.H.S. hence proved that,

$\dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} = 3.$