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If a/(b+c)+b/(c+a)+c/(a+b)=1,then a2/(b+c)+b2/(c+a)+c2/(a+b)=?
Dear Student,Please find below the solution to your problem.We know thata/(b+c) + b/(c+a) + c/(a+b) −1 = 0 which is equivalent toa^3 + b^3 + abc + c^3 / (a+b)(a+c)(b+c) = 0also= a^2/(b+c) + b^2/(c+a) + c^2/(a+b)= a^4 + a^3b + ab^3 + b^4 + a^3c + a^2bc + ab^2c + b^3c + abc^2 + ac^3 + bc^3 + c^4 /(a+b)(a+c)(b+c) = 0buta^4 + a^3b + ab^3 + b^4 + a^3c + a^2bc + ab^2c + b^3c + abc^2 + ac^3 + bc^3 + c^4 = (a+b+c)(a^3 + b^3 + abc + c^3)soa^4 + a^3b + ab^3 + b^4 + a^3c + a^2bc + ab^2c + b^3c + abc^2 + ac^3 + bc^3 + c^4=0and consequently a^2/(b+c) + b^2/(c+a) + c^2/(a+b) = 0Thanks and Regards
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