Question

已知a-b=根号3+根号2,b-c=根号3-根号2,求a平方+b平方+c平方-ab-bc-ca

Answer 1

(a-b)^2+(b-c)^2+(a-c)^2

=a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2

=2(a^2+b^2+c^2-ab-bc-ca)

将a-b=√3+√2,b-c=√3-√2,a-c=(a-b)+(b-c)=2√3带入上式计算

得(a-b)^2+(b-c)^2+(a-c)^2=22

所以a^2+b^2+c^2-ab-bc-ca=11