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

(i) (b² + d² + f²)(a² + c² + e²) = (ab + cd + ef)²

(ii) $\dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf}$

(iii) $\Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big)$

(iv) (bdf)·$\Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f)$

(i) Given,

⇒ $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k$(let)

⇒ a = kb, c = kd, e = kf.

Substituting values of a,c and e in L.H.S. of (b² + d² + f²)(a² + c² + e²) = (ab + cd + ef)², we get:

$\Rightarrow (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) \\[1em] \Rightarrow (b^2 + d^2 + f^2)(k^2 b^2 + k^2 d^2 + k^2 f^2) \\[1em] \Rightarrow (b^2 + d^2 + f^2)[k^2(b^2 + d^2 + f^2)] \\[1em] \Rightarrow k^2(b^2 + d^2 + f^2)^2.$

Substituting values of a,c and e in R.H.S. of (b² + d² + f²)(a² + c² + e²) = (ab + cd + ef)², we get:

$\Rightarrow ab + cd + ef \\[1em] \Rightarrow (k b)b + (k d)d + (k f)f \\[1em] \Rightarrow k(b^2 + d^2 + f^2) \\[1em] \therefore (ab + cd + ef)^2 = k^2(b^2 + d^2 + f^2)^2.$

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

Hence, proved that (b² + d² + f²)(a² + c² + e²) = (ab + cd + ef)² .

(ii) Given,

⇒ $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k$(let)

⇒ a = kb, c = kd, e = kf.

Substituting values of a,c and e in L.H.S. of $\dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf}$, we get:

$\Rightarrow \dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow \dfrac{k^3 b^3 + k^3 d^3 + k^3 f^3}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow \dfrac{k^3(b^3 + d^3 + f^3)}{b^{3} + d^{3} + f^{3}} \\[1em] \Rightarrow k^3.$

SSubstituting values of a,c and e in R.H.S. of $\dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf}$, we get:

$\Rightarrow \dfrac{ace}{bdf} \\[1em] \Rightarrow \dfrac{(k b)(k d)(k f)}{b d f} \\[1em] \Rightarrow k^3.$

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

Hence, proved that $\dfrac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \dfrac{ace}{bdf}$.

(iii) Given,

⇒ $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k$(let)

⇒ a = kb, c = kd, e = kf.

Substituting values of a,c and e in L.H.S. of $\Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big)$, we get:

$\Rightarrow \dfrac{a^2}{b^2} + \dfrac{c^2}{d^2} + \dfrac{e^2}{f^2} \\[1em] \Rightarrow \dfrac{k^2 b^2}{b^2} + \dfrac{k^2 d^2}{d^2} + \dfrac{k^2 f^2}{f^2} \\[1em] \Rightarrow k^2 + k^2 + k^2 \\[1em] \Rightarrow 3k^2.$

Substituting values of a,c and e in R.H.S. of $\Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big)$, we get:

$\Rightarrow \dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf} \\[1em] \Rightarrow \dfrac{(k b)(k d)}{b d} + \dfrac{(k d)(k f)}{d f} + \dfrac{(k b)(k f)}{b f} \\[1em] \Rightarrow k^2 + k^2 + k^2 \\[1em] \Rightarrow 3k^2.$

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

Hence, proved that $\Big(\dfrac{a^{2}}{b^{2}} + \dfrac{c^{2}}{d^{2}} + \dfrac{e^{2}}{f^{2}}\Big) = \Big(\dfrac{ac}{bd} + \dfrac{ce}{df} + \dfrac{ae}{bf}\Big)$.

(iv) Given,

⇒ $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k$(let)

⇒ a = kb, c = kd, e = kf.

Substituting values of a, c and d in L.H.S. of (bdf)·$\Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f)$, we get:

$\Rightarrow bdf \cdot \Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} \\[1em] \Rightarrow bdf \cdot \Big(\dfrac{kb + b}{b} + \dfrac{kd + d}{d} + \dfrac{kf + f}{f}\Big)^3 \\[1em] \Rightarrow bdf \cdot [(k + 1) + (k + 1) + (k + 1)] \\[1em] \Rightarrow bdf \cdot [3(k + 1)]^3 \\[1em] \Rightarrow bdf\cdot 27(k + 1)^3.$

Substituting values of a, c and d in R.H.S. of (bdf)·$\Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f)$, we get:

$\Rightarrow 27(a + b)(c + d)(e + f) \\[1em] \Rightarrow 27(kb + b)(kd + d)(kf + f) \\[1em] \Rightarrow 27\big(b(k + 1)\big)\big(d(k + 1)\big)\big(f(k + 1)\big) \\[1em] \Rightarrow 27 bdf(k + 1)^3.$

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

Hence, proved that (bdf).$\Big(\dfrac{a + b}{b} + \dfrac{c + d}{d} + \dfrac{e + f}{f}\Big)^{3} = 27(a + b)(c + d)(e + f)$.