Question Number 217764 by ArshadS last updated on 20/Mar/25
$$ \\ $$$$\mathrm{Let}\:\mathrm{a},\:\mathrm{b},\:\mathrm{c}\:\mathrm{be}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{a}}{\left({b}−{c}\right)^{\mathrm{2}} }+\frac{{b}}{\left({c}−{a}\right)^{\mathrm{2}} }+\frac{{c}}{\left({a}−{b}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$
Answered by vnm last updated on 20/Mar/25
$${Let}\:{P}\left({a},{b},{c}\right)={a}\left({c}−{a}\right)\left({a}−{b}\right)+{b}\left({a}−{b}\right)\left({b}−{c}\right)+{c}\left({b}−{c}\right)\left({c}−{a}\right) \\ $$$${and}\:{Q}\left({a},{b},{c}\right)={a}\left({c}−{a}\right)^{\mathrm{2}} \left({a}−{b}\right)^{\mathrm{2}} +{b}\left({a}−{b}\right)^{\mathrm{2}} \left({b}−{c}\right)^{\mathrm{2}} +{c}\left({b}−{c}\right)^{\mathrm{2}} \left({c}−{a}\right)^{\mathrm{2}} . \\ $$$${Then}\:{Q}\left({a},{b},{c}\right)={P}\left({a},{b},{c}\right)\left(−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +{bc}+{ca}+{ab}\right). \\ $$$${It}'{s}\:{not}\:{obvious},\:{of}\:{course},\:{but}\:{can}\:{be}\:{checked}. \\ $$$${So}\:{if}\:{P}\left({a},{b},{c}\right)=\mathrm{0}\:{then}\:{Q}\left({a},{b},{c}\right)=\mathrm{0}. \\ $$$$ \\ $$
Commented by ArshadS last updated on 22/Mar/25
$${Thanks}\:{a}\:{lot}! \\ $$