InterviewSolution
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InterviewSolution
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If ab+bc+ca=0, then find 1/a2-bc +1/b2-ac +1/c2 - ab |
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Answer» Given: ab+bc+ca=0To find : Value of1/(a^2-bc ) + 1/(c^2-ab) + 1/(a^2-bc)substitute -ac = ab + bc ; -ab = ac+ bc and -bc =ab + ac 1/(a^2-bc) +1/(c^2-ab) +1/(a^2-bc) =(1/(a2+ab+ac))+(1/(c2+ac+bc))+(1/(a2+ab+ac))=(1/a(a+b+c))+(1/b(a+b+c))+(1/c(a+b+c))=(ab + bc + ca)/[(a+b+c)(abc)]=0since ab+bc+ca=0 |
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