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Question Number 30456 by daffa123 last updated on 22/Feb/18

proof that (a^2 /((a−b)(a−c)))+(b^2 /((b−c)(b−a)))+(c^2 /((c−a)(c−b)))= a+b+c

$${proof}\:{that} \\ $$$$\frac{{a}^{\mathrm{2}} }{\left({a}−{b}\right)\left({a}−{c}\right)}+\frac{{b}^{\mathrm{2}} }{\left({b}−{c}\right)\left({b}−{a}\right)}+\frac{{c}^{\mathrm{2}} }{\left({c}−{a}\right)\left({c}−{b}\right)}=\:{a}+{b}+{c} \\ $$

Answered by Rasheed.Sindhi last updated on 22/Feb/18

(a^2 /((a−b)(a−c)))+(b^2 /((b−c)(b−a)))+(c^2 /((c−a)(c−b)))= a+b+c LHS ((−a^2 (b−c)−b^2 (c−a)−c^2 (a−b))/((a−b)(b−c)(c−a))) =−((a^2 b−ab^2 +b^2 c−a^2 c−bc^2 +ac^2 )/((a−b)(b−c)(c−a))) =−((ab(a−b)+c(b^2 −a^2 )−c^2 (b−a))/((a−b)(b−c)(c−a))) =−((ab(a−b)−c(a^2 −b^2 )+c^2 (a−b))/((a−b)(b−c)(c−a))) =−(((a−b){ab−c(a+b)+c^2 })/((a−b)(b−c)(c−a))) =−(((a−b){ab−ac−bc+c^2 })/((a−b)(b−c)(c−a))) =−((a(b−c)−c(b−c))/((b−c)(c−a))) =−(((b−c)(a−c))/((b−c)(c−a))) =(((b−c)(c−a))/((b−c)(c−a))) =1≠a+b+c

$$\frac{{a}^{\mathrm{2}} }{\left({a}−{b}\right)\left({a}−{c}\right)}+\frac{{b}^{\mathrm{2}} }{\left({b}−{c}\right)\left({b}−{a}\right)}+\frac{{c}^{\mathrm{2}} }{\left({c}−{a}\right)\left({c}−{b}\right)}=\:{a}+{b}+{c} \\ $$$$\mathrm{LHS} \\ $$$$\frac{−{a}^{\mathrm{2}} \left({b}−{c}\right)−{b}^{\mathrm{2}} \left({c}−{a}\right)−{c}^{\mathrm{2}} \left({a}−{b}\right)}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{{a}^{\mathrm{2}} {b}−{ab}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}−{a}^{\mathrm{2}} {c}−{bc}^{\mathrm{2}} +{ac}^{\mathrm{2}} }{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{{ab}\left({a}−{b}\right)+{c}\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)−{c}^{\mathrm{2}} \left({b}−{a}\right)}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{{ab}\left({a}−{b}\right)−{c}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} \left({a}−{b}\right)}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{\left({a}−{b}\right)\left\{{ab}−{c}\left({a}+{b}\right)+{c}^{\mathrm{2}} \right\}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{\left({a}−{b}\right)\left\{{ab}−{ac}−{bc}+{c}^{\mathrm{2}} \right\}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{{a}\left({b}−{c}\right)−{c}\left({b}−{c}\right)}{\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{\left({b}−{c}\right)\left({a}−{c}\right)}{\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=\frac{\left({b}−{c}\right)\left({c}−{a}\right)}{\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=\mathrm{1}\neq{a}+{b}+{c} \\ $$

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