Question Number 30456 by daffa123 last updated on 22/Feb/18
$${proof}\:{that} \\ $$$$\frac{{a}^{\mathrm{2}} }{\left({a}−{b}\right)\left({a}−{c}\right)}+\frac{{b}^{\mathrm{2}} }{\left({b}−{c}\right)\left({b}−{a}\right)}+\frac{{c}^{\mathrm{2}} }{\left({c}−{a}\right)\left({c}−{b}\right)}=\:{a}+{b}+{c} \\ $$
Answered by Rasheed.Sindhi last updated on 22/Feb/18
$$\frac{{a}^{\mathrm{2}} }{\left({a}−{b}\right)\left({a}−{c}\right)}+\frac{{b}^{\mathrm{2}} }{\left({b}−{c}\right)\left({b}−{a}\right)}+\frac{{c}^{\mathrm{2}} }{\left({c}−{a}\right)\left({c}−{b}\right)}=\:{a}+{b}+{c} \\ $$$$\mathrm{LHS} \\ $$$$\frac{−{a}^{\mathrm{2}} \left({b}−{c}\right)−{b}^{\mathrm{2}} \left({c}−{a}\right)−{c}^{\mathrm{2}} \left({a}−{b}\right)}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{{a}^{\mathrm{2}} {b}−{ab}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}−{a}^{\mathrm{2}} {c}−{bc}^{\mathrm{2}} +{ac}^{\mathrm{2}} }{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{{ab}\left({a}−{b}\right)+{c}\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)−{c}^{\mathrm{2}} \left({b}−{a}\right)}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{{ab}\left({a}−{b}\right)−{c}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} \left({a}−{b}\right)}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{\left({a}−{b}\right)\left\{{ab}−{c}\left({a}+{b}\right)+{c}^{\mathrm{2}} \right\}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{\left({a}−{b}\right)\left\{{ab}−{ac}−{bc}+{c}^{\mathrm{2}} \right\}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{{a}\left({b}−{c}\right)−{c}\left({b}−{c}\right)}{\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=−\frac{\left({b}−{c}\right)\left({a}−{c}\right)}{\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=\frac{\left({b}−{c}\right)\left({c}−{a}\right)}{\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$=\mathrm{1}\neq{a}+{b}+{c} \\ $$