a+b+c=0,1/（a+1）+1/（b+2）+1/（c+3）=0,求（a+1）的平方+（b+2）的平方+（c+3）的平

a+b+c=0,(a+1)+(b+2)+(c+3)=6,
1/(a+1)+1/(b+2)+1/(c+3)=0,得(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)=0
（a+1)^2+（b+2)^2+（c+3)^2
＝〔(a+1)+(b+2)+(c+3)〕＾2-2〔(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)〕
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