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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,aa2−bc+bb2−ca+cc2−ab=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=0⇒a2=−ab−cab2=−bc−ab,c2=−ca−bcaa2−bc+bb2−ca+cc2−ab=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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