已知实数 a,b,c 满足{abc=−1,a+b+c=4,aa2−3a−1+bb2−ba−1+cc2−3c−1=49,
则 a2+b2+c2= _______．
答案 332．
解析 由于 abc=−1，a+b+c=4，有a2−3a−1=a2−3a+abc=a(bc+a−3)=a(bc−b−c+1)=a(b−1)(c−1),
于是aa2−3a−1=1(b−1)(c−1),
同理可得：bb2−3b−1=1(a−1)(c−1),cc2−3c−1=1(a−1)(b−1),
又aa2−3a−1+bb2−3b−1+cc2−3c−1=49,
因此1(b−1)(c−1)+1(a−1)(c−1)+1(a−1)(b−1)=49,
从而(a−1)+(b−1)+(c−1)(a−1)(b−1)(c−1)=49,
即49(a−1)(b−1)(c−1)=(a−1)+(b−1)+(c−1).
整理得49(abc−ab−ac−bc+a+b+c−1)=a+b+c−3,
将 abc=−1，a+b+c=4 代入得ab+bc+ac=−14,
则a2+b2+c2=(a+b+c)2−2(ab+bc+ac)=332.