Question

If x/(b-c)(b+c) = y/(c-a)(c+a) = z/(a−b)(a+b) prove that x + y + z =0​

Answer 1

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

Given that,

$\sf \: \dfrac{x}{(b - c)(b + c)} = \dfrac{y}{(c - a)(c + a)} = \dfrac{z}{(a - b)(a + b)}$

$\sf \: Let \: \: \dfrac{x}{ {b}^{2} - {c}^{2} } = \dfrac{y}{ {c}^{2} - {a}^{2} } = \dfrac{z}{ {a}^{2} - {b}^{2} } = k$

$\: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2}}$

$\red{\rm :\longmapsto\:x = k( {b}^{2} - {c}^{2}) - - - (1)} \\ \red{\rm :\longmapsto\:y = k( {c}^{2} - {a}^{2}) - - - (2)} \\ \red{\rm :\longmapsto\:z = k( {a}^{2} - {b}^{2}) - - - (3)}$

On adding equation (1),(2) and (3), we get

$\rm :\longmapsto\:x + y + z = k( \red{ \cancel {b}^{2} } - \blue{ \cancel {c}^{2}} + \blue{ \cancel {c}^{2} } - \green{ \cancel {a}^{2} }+ \green{ \cancel {a}^{2} } - \red{ \cancel {b}^{2} })$

$\bf\implies \:x + y + z = 0$

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

Additional Information

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

$\boxed{ \sf \: \dfrac{a}{c} = \dfrac{b}{d} \: \: \: (alternendo)}$

$\boxed{ \sf \: \dfrac{b}{a} = \dfrac{d}{c} \: \: \: (invertendo)}$

$\boxed{ \sf \: \dfrac{a + b}{b} = \dfrac{c + d}{d} \: \: \: (componendo)}$

$\boxed{ \sf \: \dfrac{a - b}{b} = \dfrac{c - d}{d} \: \: \: (dividendo)}$

$\boxed{ \sf \: \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{a + c}{b + d} \: \: \: (addendo)}$