Question

$\mathfrak{ \large{ \red{ \underbrace{Question}}}}$
$\sf{ \bold{q.(1)} \: if \: x = \frac{4ab}{a + b} \: \: show \: that \ratio - } \\ \\ \sf{ \bold{\frac{x + 2a}{x - 2a} + \frac{x + 2b}{x - 2b} }}$
$\sf{ \bold{q.(2)}} \: if \: x = \frac{ \sqrt[3]{m + 1 } + \sqrt[3]{m - 1} }{ \sqrt[3]{m + 1} - \sqrt[3]{m - 1} } \: prove \: that \ratio - \\ \\ \sf{ \bold{ \: {x}^{3} - 3m {x}^{2} + 3x - m = 0}}$

$\large{ \red{ \sf{note}}}$

$\rightarrow {\rm{proper \: answer \: with \: explanation \: needed}}$
$\rightarrow{ \rm{spam = 10 \: answer \: report}}$
​

Answer 1

Basic Concept Used :-

$\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d} \: then$

$\boxed{ \bf \: \dfrac{a + b}{b} = \dfrac{c + d}{d} \: \: \: is \: called \: componendo}$

$\boxed{ \bf \: \dfrac{a - b}{b} = \dfrac{c - d}{d} \: \: \: is \: called \: dividendo}$

$\boxed{ \bf \: \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}\: is \: called \: componendo \: and \: dividendo}$

$\large\underline{\sf{Solution-1}}$

Given that

$\rm :\longmapsto\:x = \dfrac{4ab}{a + b}$

$\rm :\longmapsto\:x = \dfrac{2a \times 2b}{a + b}$

$\rm :\longmapsto\:\dfrac{x}{2a} = \dfrac{2b}{a + b}$

Apply Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{x + 2a}{x - 2a} = \dfrac{2b + a + b}{2b - (a + b)}$

$\rm :\longmapsto\:\dfrac{x + 2a}{x - 2a} = \dfrac{3b + a}{b - a} - - - (1)$

Again,

We have

$\rm :\longmapsto\:x = \dfrac{4ab}{a + b}$

$\rm :\longmapsto\:x = \dfrac{2a \times 2b}{a + b}$

$\rm :\longmapsto\:\dfrac{x}{2b} = \dfrac{2a}{a + b}$

On applying Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{x + 2b}{x - 2b} = \dfrac{2a + a + b}{2a - (a + b)}$

$\rm :\longmapsto\:\dfrac{x + 2b}{x - 2b} = \dfrac{3a+ b}{a -b} - - - (2)$

On adding equation (1) and equation (2), we get

$\rm :\longmapsto\:\dfrac{x + 2a}{x - 2a} + \dfrac{x + 2b}{x - 2b} =\dfrac{3b + a}{b - a} + \dfrac{3a+ b}{a -b}$

$\rm :\longmapsto\:\dfrac{x + 2a}{x - 2a} + \dfrac{x + 2b}{x - 2b} =\dfrac{3b + a}{b - a} - \dfrac{3a+ b}{b - a}$

$\rm :\longmapsto\:\dfrac{x + 2a}{x - 2a} + \dfrac{x + 2b}{x - 2b} =\dfrac{3b + a - 3a - b}{b - a}$

$\rm :\longmapsto\:\dfrac{x + 2a}{x - 2a} + \dfrac{x + 2b}{x - 2b} =\dfrac{2b - 2a}{b - a}$

$\rm :\longmapsto\:\dfrac{x + 2a}{x - 2a} + \dfrac{x + 2b}{x - 2b} =\dfrac{2 \: \: \cancel{(b - a)}}{\cancel{b - a}}$

$\bf\implies \:\:\dfrac{x + 2a}{x - 2a} + \dfrac{x + 2b}{x - 2b} =2$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

$\large\underline{\sf{Solution-2}}$

$\rm\: x = \dfrac{ \sqrt[3]{m + 1 } + \sqrt[3]{m - 1} }{ \sqrt[3]{m + 1} - \sqrt[3]{m - 1} }$

can be rewritten as,

$\rm :\longmapsto\:\dfrac{x}{1} = \dfrac{ \sqrt[3]{m + 1 } + \sqrt[3]{m - 1} }{ \sqrt[3]{m + 1} - \sqrt[3]{m - 1} }$

On applying Componendo and Dividendo, we get

$\rm \:\dfrac{x+1}{x-1}= \dfrac{\sqrt[3]{m+1}+\cancel{\sqrt[3]{m-1}}+ \sqrt[3]{m + 1}-\cancel{\sqrt[3]{m - 1}}}{\cancel{\sqrt[3]{m + 1}} + \sqrt[3]{m - 1}-\cancel{\sqrt[3]{m + 1}}+\sqrt[3]{m - 1}}$

$\rm \:\dfrac{x+1}{x-1}= \dfrac{\cancel{2} \: \sqrt[3]{m+1}}{\cancel{2} \: \sqrt[3]{m - 1}}$

On cubing both sides, we get

$\rm :\longmapsto\:\dfrac{ {(x + 1)}^{3} }{ {(x - 1)}^{3} } = \dfrac{m + 1}{m - 1}$

$\rm :\longmapsto\:\dfrac{ {x}^{3} + {1}^{3} + 3x(x + 1) }{ {x}^{3} - {(1)}^{3} - 3x(x - 1)} = \dfrac{m + 1}{m - 1}$

$\bigg \{\pink{\rm\because\: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)} \\ \pink{\rm\because\: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)} \bigg \}$

$\rm :\longmapsto\:\dfrac{ {x}^{3} +1+ {3x}^{2} + 3x}{ {x}^{3} -1 - {3x}^{2} + 3x } = \dfrac{m + 1}{m - 1}$

$\rm :\longmapsto\:\dfrac{ ({x}^{3} + {3x}) + ( {3x}^{2} + 1) }{( {x}^{3} + {3x}) - ( {3x}^{2} + 1)} = \dfrac{m + 1}{m - 1}$

On applying Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{ {x}^{3} + {3x}}{ {3x}^{2} + 1} = \dfrac{m}{1}$

$\rm :\longmapsto\: {x}^{3} + {3x} = 3m {x}^{2} + m$

$\bf\implies \: {x}^{3} - {3mx}^{2} + 3x - m = 0$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d} \: then$

$\boxed{ \bf \: \dfrac{a}{c} = \dfrac{b}{d} \: is \: called \: alternendo}$

$\boxed{ \bf \: \dfrac{b}{a} = \dfrac{d}{c} \: is \: called \: invertendo}$