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Question Number 191615 by MATHEMATICSAM last updated on 27/Apr/23
a + b + c = 0. Prove that,  (a/(a^2  − bc)) + (b/(b^2  − ca)) + (c/(c^2  − ab)) = 0.
a+b+c=0.Provethat,aa2bc+bb2ca+cc2ab=0.
Answered by som(math1967) last updated on 27/Apr/23
 a+b+c=0  ⇒a^2 =−ab−ca  b^2 =−bc−ab, c^2 =−ca−bc  (a/(a^2 −bc))+(b/(b^2 −ca))+(c/(c^2 −ab))  =(a/(−(ab+bc+ca)))+(b/(−(ab+bc+ca)))    +(c/(−(ab+bc+ca)))  =((a+b+c)/(−(ab+bc+ca)))=(0/(−(ab+bc+ca)))=0
a+b+c=0a2=abcab2=bcab,c2=cabcaa2bc+bb2ca+cc2ab=a(ab+bc+ca)+b(ab+bc+ca)+c(ab+bc+ca)=a+b+c(ab+bc+ca)=0(ab+bc+ca)=0

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