Question

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

Answer 1

相加
a-c=4

所以原式=(2a²+2b²+2c²-2ab-2bc-2ac)/2
=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)]/2
=[(a-b)²+(b-c)²+(a-c)²]/2
=(7+4√3+7-4√3+16)/2
=15

Answer 2

a-b=2+√3,b-c=2-√3,
=> a-c=a-b+b-c=4, a+b-2c=(a-b)+2(b-c)=2+√3+2(2-√3)=6-√3,
则
a^2+b^2+c^2-ab-bc-ca
=a(a-b)+b(b-c)+c(c-a)
=a(2+√3)+b(2-√3)-4c
=2(a+b)+√3(a-b)-4c=2(a+b-2c)+√3(a-b)
=2(6-√3)+√3(2+√3)
=15