答案
解:由题意可知a-b=-1,b-c=-1,a-c=-2
所求式=
(2a
2+2b
2+2c
2-2ab-2bc-2ca)
=
[(a
2-2ab+b
2)+(b
2-2bc+c
2)+(a
2-2ac+c
2)]
=
[(a-b)
2+(b-c)
2+(a-c)
2]
=
[(-1)
2+(-1)
2+(-2)
2]=3.
解:由题意可知a-b=-1,b-c=-1,a-c=-2
所求式=
(2a
2+2b
2+2c
2-2ab-2bc-2ca)
=
[(a
2-2ab+b
2)+(b
2-2bc+c
2)+(a
2-2ac+c
2)]
=
[(a-b)
2+(b-c)
2+(a-c)
2]
=
[(-1)
2+(-1)
2+(-2)
2]=3.