If $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f}$, prove that each ratio is equal to

(i) $\sqrt{\dfrac{3a^2 - 5c^2 + 7e^2}{3b^2 - 5d^2 + 7f^2}}$

(ii) $\Big(\dfrac{2a^3 + 5c^3 + 7e^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3}.$

(i) Let, $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k,$ then,

a = bk, c = dk, e = fk.

Given,

$\sqrt{\dfrac{3a^2 - 5c^2 + 7e^2}{3b^2 - 5d^2 + 7f^2}}$

Putting values of a, c, e in above equation,

$= \sqrt{\dfrac{3(bk)^2 - 5(dk)^2 + 7(fk)^2}{3b^2 - 5d^2 + 7f^2}} \\[1em] = \sqrt{\dfrac{3b^2k^2 - 5d^2k^2 + 7f^2k^2}{3b^2 - 5d^2 + 7f^2}} \\[1em] = \sqrt{k^2\Big(\dfrac{3b^2 - 5d^2 + 7f^2}{3b^2 - 5d^2 + 7f^2}\Big)} \\[1em] = k.$

Since, the values of all ratios = k. Hence, proved.

(ii) Let, $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k,$ then,

a = bk, c = dk, e = fk.

Given,

$\Big(\dfrac{2a^3 + 5c^3 + 7e^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3}$

Putting values of a, c, e in above equation, $= \Big(\dfrac{2(bk)^3 + 5(dk)^3 + 7(fk)^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3} \\[1em] = \Big(\dfrac{2b^3k^3 + 5d^3k^3 + 7f^3k^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3} \\[1em] = \Big[k^3\Big(\dfrac{2b^3 + 5d^3 + 7f^3}{2b^3 + 5d^3 + 7f^3}\Big)\Big]^{1/3} \\[1em] = k^{3 \times 1/3} \\[1em] = k.$

Since, the values of all ratios = k. Hence, proved.