Question

已知a-b=3,b-c=-2,求a^2+b^2+c^2-ab-bc-ca (多种方法，至少三种)

Answer 1

一、a-b=3,b-c=-2,
a-c=3+(-2)=1
a^2+b^2+c^2-ab-bc-ca
=1/2(2a^2+2b^2+2c^2-2ab-2bc-2ca)
=1/2[(a-b)^2+(b-c)^2+(c-a)^2]
=1/2(3^2+2^2+1^2)
=7
二、c=b+2
a^2+b^2+c^2-ab-bc-ca
=a^2+b^2+(b+2)^2-ab-b(b+2)-(b+2)a
=a^2+b^2-2ab+2b-2a+4
=(a-b)^2-2(a-b)+4
=3^2-2*3+4
=7
三、a^2+b^2-2ab=9,a^2+b^2=2ab+9
a^2+b^2+c^2-ab-bc-ca
=2ab+9+c^2-ab-bc-ca
=9+ab-ca-bc+c^2
=9+a(b-c)-c(b-c)
=9+(a-c)(b-c)
=9+1*(-2)
=7

Answer 2

a-b=3 b-c=-2
∴a-c=(a-b)+(b-c)=3+(-2)=1
a²+b²+c²-ab-bc-ca
=1/2[(a-b)²+(b-c)²+(c-a)²]
=1/2[9+4+1]
=7