8. Suppose that $a, b, c, t$ satisfy all the conditions. Then $a b c \neq 0$ and
$$x_{1} x_{2}=\frac{c}{a}, \quad x_{2} x_{3}=\frac{a}{b}, \quad x_{3} x_{1}=\frac{b}{c} .$$
Multiplying these equations, we obtain $x_{1}^{2} x_{2}^{2} x_{3}^{2}=1$, and hence $x_{1} x_{2} x_{3}=\varepsilon= \pm 1$. From (1) we get $x_{1}=\varepsilon b / a, x_{2}=\varepsilon c / b, x_{3}=\varepsilon a / c$. Substituting $x_{1}$ in the first equation, we get $a b^{2} / a^{2}+t \varepsilon b^{2} / a+c=0$, which gives us
$$b^{2}(1+t \varepsilon)=-a c .$$
Analogously, $c^{2}(1+t \varepsilon)=-a b$ and $a^{2}(1+t \varepsilon)=-b c$, and therefore $(1+t \varepsilon)^{3}=-1$; i.e., $1+t \varepsilon=-1$, since it is real. This also implies together with (1) that $b^{2}=a c, c^{2}=a b$, and $a^{2}=b c$, and consequently
$$a=b=c \text {. }$$
Thus the three equations in the problem are equal, which is impossible. Hence, such $a, b, c, t$ do not exist.