$bx_2^2+cx_2+a=0$ and $ax_2^3+bx_2^2+cx_2=0 \to x_2^3=1$ Same for $x_1,x_3$ So $x_1,x_2,x_3$ are different roots of $x^3-1=0$ so let $x_1=1$ But then $a+b+c=0$ and so $x=1$ is common root for all equations, so contradiction

Suppose instead that there do exist such $a,b,c$. Now, from vieta's formulas we get: \begin{align*} &x_1 + x_2 = -\frac ba && x_1x_2 = \frac ca \\ &x_2 + x_3 = -\frac cb && x_2x_3 = \frac ab \\ &x_3 + x_1 = -\frac ac && x_1x_2 = \frac bc \end{align*}Thus, $$(x_1x_2x_3)^2 = (x_1x_2)(x_2x_3)(x_3x_1) = \frac ca \cdot \frac ab \cdot \frac bc = 1 \implies x_1x_2x_3 = \pm 1.$$Also, $$(x_1 + x_2)x_2x_3 = -\frac ba \cdot \frac ab = -1 \implies x_2^2x_3 = -1 - x_1x_2x_3.$$Note that $a,b,c$ being non zero implies that $x_1,x_2,x_3$ are also non zero. So, $x_1x_2x_3$ cannot be $-1$. Thus, $x_1x_2x_3 = 1$, and so $x_2^2x_3 = -2$. Thus, $x_2 = -2x_1$. Similarly, $x_3 = -2x_2$ and $x_1 = -2x_2$. Therefore, $x_1 = -8x_1$ and hence $x_1 = 0$ which is a contradiction. $\blacksquare$