For all admissible a, b, c find all possible values of the expression (a−c)(b−c)(a+b−c)2+(b−a)(c−a)(b+c−a)2+(c−b)(a−b)(c+a−b)2
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Set A=a+b+c. Then M=(a−c)(b−c)(a+b−c)2+(b−a)(c−a)(b+c−a)2+(c−b)(a−b)(c+a−b)2==(a−c)(b−c)(A−2c)2+(b−a)(c−a)(A−2a)2+(c−b)(a−b)(A−2b)2==(a−b)(b−c)(c−a)(A−2c)2(b−a)+(A−2a)2(c−b)+(A−2b)2(a−c)=(a−b)(b−c)(c−a)L
We have L=(A2−4Ac+4c2)(b−a)+(A2−4Aa+4a2)(c−b)+(A2−4Ab+4b2)(a−c)=A2((b−a)+(c−b)+(a−c))−−2A(c(b−a)+a(c−b)+b(a−c))+4(c2(b−a)+a2(c−b)+b2(a−c))==−4A(cb−ca+ac−ab+ba−bc)+4N=4N.
Further N=c2(b−a)+a2(c−b)+b2(a−c)=c2(b−a)+(a2c−a2b+b2a−b2c)==c2(b−a)+(a2c−b2c)−(a2b−b2a)=c2(b−a)+c(a+b)(a−b)−ab(a−b)==(a−b)(ca+cb−ab−c2)=(a−b)(a(c−b)+c(b−c))=(a−b)(b−c)(c−a). Therefore, M=4.