问题
解答题
已知a+b+c=0且abc≠0，求
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答案
∵a+b+c=0，即a=-（b+c），b+c=-a，
∴原式=
+1 (b+c)2-b2-c2
+1 b2-(b+c)2-c2 1 c2-(b+c)2-b2
=
+1 2bc
+1 -2bc-2c2
=1 -2bc-2b2
-1 2bc
-1 2c(b+c) 1 2b(c+b)
=
+1 2bc
+1 2ac
=1 2ab
=0．a+b+c 2abc