We have given the following system of equations: $(1) \;\; x^3 - ax^2 + b^3 = 0$ $(2) \;\; x^3 - bx^2 + c^3 = 0$ $(3) \;\; x^3 - cx^2 + a^3 = 0$ Assume that the $a$, $b$ and $c$ are different. We may WLOG (since $a$, $b$ and $c$ are symmetric in (1)-(3)) assume $a = \max\{a,b,c\}$. By subtracting equation (1) from equation (3), the result is $(a - c)x^2 + a^3 - b^3 = 0$, yielding (since $a$, $b$ and $c$ are distinct), $(4) \;\; x^2 = \frac{b^3 - a^3}{a - c}$. Now $a = \max\{a,b,c\}$ means $a - c > 0$ and $b^3 - a^3 < 0$, implying ${\textstyle \frac{b^3 - a^3}{a - c} < 0}$. Hence according to equation (4) we obtain $x^2 < 0$, which obviously is impossible. In other words, the system of equations (1)-(3) is not solvable if $a$, $b$ and $c$ are different. q.e.d. Comment: To find a criteria for solvability of the system of equations (1)-(3), we can use the fact that a solution exists only if $a$, $b$ and $c$ are not different. Assume $a=c$. Then we know from the solution above that $(a - c)x^2 = b^3 - a^3$, yielding $b^3 = a^3$ (since $a=c$), which implies $b=a$. Therefore the system of equations (1)-(3) is solvable only if $a=b=c$.