Question

已知a-b=根号3+根号2,b-c=根号3-根号2。求a²+b²+c²-ab-bc-ca的值

如题，要有解答过程，详细点，谢谢。

Answer 1

a-b=√3+√2 1
b-c=√3-√2 2
1式+2式得
a-c=2√3
a²+b²+c²-ab-bc-ca
=1/2*(2a^2+2b^2+2c^2-2ab-2bc-2ca)
=1/2(a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2)
=1/2[(a-b)^2+(a-c)^2+(b-c)^2]
=1/2[(√3+√2)^2+(2√3)^2+(√3-√2)^2]
=1/2(3+2√6+2+12+3-2√6+2)
=1/2*22
=11