已知实数 $a,b,c$ 满足\[\begin{cases} abc=-1,\\ a+b+c=4,\\ \dfrac {a}{a^2-3a-1}+\dfrac {b}{b^2-ba-1}+\dfrac {c}{c^2-3c-1}=\dfrac 4 9,\end{cases}\]则 $a^2+b^2+c^2=$ _______．
答案 $\dfrac {33}{2}$．
解析 由于 $abc=-1$，$a+b+c=4$，有\[a^2-3a-1=a^2-3a+abc=a(bc+a-3)=a(bc-b-c+1)=a(b-1)(c-1),\]于是\[\dfrac {a}{a^2-3a-1}=\dfrac {1}{(b-1)(c-1)},\]同理可得：\[\dfrac {b}{b^2-3b-1}=\dfrac {1}{(a-1)(c-1)} , \dfrac {c}{c^2-3c-1}=\dfrac {1}{(a-1)(b-1)},\]又\[\dfrac {a}{a^2-3a-1}+\dfrac {b}{b^2-3b-1}+\dfrac {c}{c^2-3c-1}=\dfrac 4 9,\]因此\[ \dfrac {1}{(b-1)(c-1)}+\dfrac {1}{(a-1)(c-1)}+\dfrac {1}{(a-1)(b-1)}=\dfrac 4 9,\]从而\[\dfrac {(a-1)+(b-1)+(c-1)}{(a-1)(b-1)(c-1)}=\dfrac 4 9,\]即\[\dfrac 4 9(a-1)(b-1)(c-1)=(a-1)+(b-1)+(c-1).\]整理得\[\dfrac 4 9(abc-ab-ac-bc+a+b+c-1)=a+b+c-3,\]将 $abc=-1$，$a+b+c=4$ 代入得\[ab+bc+ac=-\dfrac 1 4,\]则\[a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=\dfrac {33}{2}.\]