有道难题想请教如果a+b+c=0 1/(a+1)+1/(b+2)+1/(c+3)=0求：(a+1)(a+1)+(b+2)

36
a+b+c=0,所以(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
那么：(a+1)(a+1)+(b+2)(b+2)+(c+3)(c+3)
＝〔(a+1)+(b+2)+(c+3)〕＾2－2〔(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)〕
＝36－0
＝36