We factor the first and third givens, obtaining the system $$\begin{aligned} a^{2} b c+a b^{2} c+a b c^{2}-a-b-c=(a b c-1)(a+b+c) & =-8 \\ a^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b+3 a b c=(a b+b c+c a)(a+b+c) & =-4 \\ a^{2} b^{2} c+a b^{2} c^{2}+a^{2} b c^{2}-a b-b c-c a=(a b c-1)(a b+b c+c a) & =2 \end{aligned}$$ Writing $X=a+b+c, Y=a b+b c+c a, Z=a b c-1$, we have $X Z=-8, X Y=-4, Y Z=$ 2. Multiplying the three yields $(X Y Z)^{2}=64$ from which $X Y Z= \pm 8$. Since we are given $X>0$, multiplying the last equation by $X$ we have $2 X=X Y Z= \pm 8$. Evidently $X Y Z=8$ from which $X=4, Y=-1, Z=-2$. We conclude that $a, b, c$ are the roots of the polynomial $P(t)=t^{3}-4 t^{2}-t+1$. Thus, $P(a)=a^{3}-4 a^{2}-a+1=0$, and also $P(b)=P(c)=0$. Now since $P(1 / 2)=-\frac{3}{8}, P(0)=1$ and $P(-2 / 3)=-\frac{11}{27}$, we conclude that $-2 / 3<c<0<b<1 / 2<a$. It follows that $\left|b^{5}+c^{5}\right|<\frac{1}{2}$. Thus, we compute $a^{5}+b^{5}+c^{5}$. Defining $S_{n}=a^{n}+b^{n}+c^{n}$, we have $S_{n+3}=4 S_{n+2}+S_{n+1}-S_{n}$ for $n \geq 0$. Evidently $S_{0}=3, S_{1}=4, S_{2}=(a+b+c)^{2}-2(a b+b c+c a)=18$. Then $S_{3}=4 \cdot 18+4-3=73$, $S_{4}=4 \cdot 73+18-4=306$, and $S_{5}=4 \cdot 306+73-18=1279$. Since $\left|b^{5}+c^{5}\right|<\frac{1}{2}$, we conclude that $\left|S_{5}-a^{5}\right|<\frac{1}{2}$, or that 1279 is the integer nearest to $a^{5}$.