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

$\begin{array}{ll} \text{(i)} & \dfrac{a + b}{b + c} = \dfrac{a^2(b - c)}{b^2(a - b)} \\ \text{(ii)} & \dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3} \ & = \dfrac{a}{b^2c^2} + \dfrac{b}{c^2a^2} + \dfrac{c}{a^2b^2} \\ \text{(iii)} & a : c = (a^2 + b^2) : (b^2 + c^2) \\ \text{(iv)} & a^2b^2c^2(a^{-4} + b^{-4} + c^{-4}) \ & = b^{-2}(a^4 + b^4 + c^4) \\ \text{(v)} & abc(a + b + c)^3 \ & = (ab + bc + ca)^3 \\ \text{(vi)} & (a + b + c)(a - b + c) \ & = a^2 + b^2 + c^2. \end{array}$

(i) Since, a, b, c are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = \dfrac{a + b}{b + c} \\[1em] = \dfrac{ck^2 + ck}{ck + c} \\[1em] = \dfrac{ck(k + 1)}{c(k + 1)} \\[1em] = k. \\[1em] \text{R.H.S.} = \dfrac{a^2(b - c)}{b^2(a - b)} \\[1em] = \dfrac{(ck^2)^2(ck - c)}{(ck)^2(ck^2 - ck)} \\[1em] = \dfrac{c^2k^4c(k - 1)}{c^2k^2ck(k - 1)} \\[1em] = \dfrac{c^3k^4(k - 1)}{c^3k^3(k - 1)} \\[1em] = k.$

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

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

(ii) Since, a, b, c are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = \dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3} \\[1em] = \dfrac{1}{(ck^2)^3} + \dfrac{1}{(ck)^3} + \dfrac{1}{c^3} \\[1em] = \dfrac{1}{c^3k^6} + \dfrac{1}{c^3k^3} + \dfrac{1}{c^3} \\[1em] = \dfrac{1}{c^3}\Big(\dfrac{1}{k^6} + \dfrac{1}{k^3} + 1\Big) \\[1em] \text{R.H.S.} = \dfrac{a}{b^2c^2} + \dfrac{b}{c^2a^2} + \dfrac{c}{a^2b^2} \\[1em] = \dfrac{ck^2}{(ck)^2c^2} + \dfrac{ck}{c^2(ck^2)^2} + \dfrac{c}{(ck^2)^2(ck)^2} \\[1em] = \dfrac{ck^2}{c^4k^2} + \dfrac{ck}{c^4k^4} + \dfrac{c}{c^4k^6} \\[1em] = \dfrac{1}{c^3} + \dfrac{1}{c^3k^3} + \dfrac{1}{c^3k^6} \\[1em] = \dfrac{1}{c^3}\Big(1 + \dfrac{1}{k^3} + \dfrac{1}{k^6}\Big) \\[1em]$

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

$\dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3} = \dfrac{a}{b^2c^2} + \dfrac{b}{c^2a^2} + \dfrac{c}{a^2b^2}$.

(iii) Since, a, b, c are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = a : c \\[1em] = \dfrac{a}{c} \\[1em] = \dfrac{ck^2}{c} \\[1em] = k^2. \\[1em] \text{R.H.S.} = (a^2 + b^2) : (b^2 + c^2) \\[1em] = \dfrac{a^2 + b^2}{b^2 + c^2} \\[1em] = \dfrac{((ck^2)^2 + (ck)^2)}{((ck)^2 + c^2)} \\[1em] = \dfrac{c^2k^4 + c^2k^2}{c^2k^2 + c^2} \\[1em] = \dfrac{c^2k^2(k^2 + 1)}{c^2(k^2 + 1)} \\[1em] = k^2. \\[1em]$

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

a : c = (a² + b²) : (b² + c²).

(iv) Since, a, b, c are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = a^2b^2c^2(a^{-4} + b^{-4} + c^{-4}) \\[1em] = (ck^2)^2(ck)^2c^2((ck^2)^{-4} + (ck)^{-4} + c^{-4}) \\[1em] = c^2k^4c^2k^2c^2(c^{-4}k^{-8} + c^{-4}k^{-4} + c^{-4}) \\[1em] = c^6k^6c^{-4}(k^{-8} + k^{-4} + 1) \\[1em] = c^2k^6\big(\dfrac{1}{k^8} + \dfrac{1}{k^4} + 1\big) \\[1em] = c^2k^6\big(\dfrac{1 + k^4 + k^8}{k^8}\big) \\[1em] = \dfrac{c^2k^6}{k^8}(1 + k^4 + k^8) \\[1em] = \dfrac{c^2}{k^2}(1 + k^4 + k^8) \\[1em] \text{R.H.S.} = b^{-2}(a^4 + b^4 + c^4) \\[1em] = (ck)^{-2}((ck^2)^4 + (ck)^4 + c^4) \\[1em] = c^{-2}k^{-2}(c^4k^8 + c^4k^4 + c^4) \\[1em] = c^{-2}k^{-2}c^4(k^8 + k^4 + 1) \\[1em] = c^2k^{-2}(k^8 + k^4 + 1) \\[1em] = \dfrac{c^2}{k^2}(1 + k^4 + k^8). \\[1em]$

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

a²b²c²(a^-4 + b^-4 + c^-4) = b^-2(a⁴ + b⁴ + c⁴).

(v) Since, a, b, c are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = abc(a + b + c)^3 \\[1em] = (ck^2)(ck)c(ck^2 + ck + c)^3 \\[1em] = c^3k^3(c^3(k^2 + k + 1)^3) \\[1em] = c^6k^3(k^2 + k + 1)^3 \\[1em] \text{R.H.S.} = (ab + bc + ca)^3 \\[1em] = ((ck^2)(ck) + (ck)(c) + (c)(ck^2))^3 \\[1em] = (c^2k^3 + c^2k + c^2k^2)^3 \\[1em] = ((c^2k)^3(k^2 + 1 + k)^3) \\[1em] = c^6k^3(k^2 + k + 1)^3. \\[1em]$

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

abc(a + b + c)³ = (ab + bc + ca)³.

(vi) Since, a, b, c are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = k \\[1em] \Rightarrow b = ck \text{ and } a = bk = ck^2 \\[1em] \text{L.H.S.} = (a + b + c)(a - b + c) \\[1em] = (ck^2 + ck + c)(ck^2 - ck + c) \\[1em] = c(k^2 + k + 1)c(k^2 - k + 1) \\[1em] = c^2(k^4 - k^3 + k^2 + k^3 - k^2 + k + k^2 - k + 1) \\[1em] = c^2(k^4 - \bcancel{k^3} + \bcancel{k^2} + \bcancel{k^3} - \bcancel{k^2} + \bcancel{k} + k^2 - \bcancel{k} + 1) \\[1em] = c^2(k^4 + k^2 + 1) \\[1em] \text{R.H.S.} = a^2 + b^2 + c^2 \\[1em] = (ck^2)^2 + (ck)^2 + c^2 \\[1em] = c^2k^4 + c^2k^2 + c^2 \\[1em] = c^2(k^4 + k^2 + 1). \\[1em]$

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

(a + b + c)(a - b + c) = a² + b² + c².