Cardinality of a set of polynomials where the sum of the cubes of the roots is zero Let $C\subseteq \mathbb Z\times \mathbb Z$ be the set of integer pairs $(a,b)$ for which the  3 complex roots $r_1,r_2,r_3$ of the polynomial $p(x)=x^3-2x^2+ax+b$ satisfy $r_1^3+r_2^3+r_3^3=0$ .Then what will be the cardinality of $ C $ ? 
 A: Applying Vieta's formulas, we know that $$\begin{align}
r_1+r_2+r_3 & = 2 \tag{A}\\
r_1r_2 + r_1r_3 + r_2r_3 & = a\tag{B} \\
r_1r_2r_3 & = -b \tag{C}
\end{align}$$
Multiplying $(A)$ by $(B)$  yields:
$$r_1^2r_2 + r_1^2r_3 + r_1r_2^2 + r_1r_3^2 + r_2^2r_3 + r_2r_3^2 +
3r_1r_2r_3 = 2a \tag{D}$$
and then subtracting three times $(C)$ we get:
$$r_1^2r_2 + r_1^2r^3 + r_1r_2^2 + r_1r_3^2 + r_2^2r_3 + r_2r_3^2 = 2a+3b.\tag{E}$$
Now if we cube $(A)$, we get $$\begin{align}
r_1^3 +r_2^3 + r_3^3 + \\
3(r_1^2r_2 + r_1^2r_3 + r_1r_2^2 + r_1r_3^2 + r_2^2r_3 + r_2r_3^2) + \\
6r_1r_2r_3 &= 8\tag{F}\end{align}$$
and by combining this with $(E)$ and $(C)$, we obtain finally
$$\begin{align}
r_1^3 +r_2^3  + r_3^3& + 3(2a+3b) + 6(-b)  &&=8\\
r_1^3 +r_2^3 + r_3^3 & && = 8-6a-3b\tag{$\ast$}
\end{align}$$
(Taking $a=b=0$ and $a=1, b=0$ provides a check that this result is correct.)
In order to have $r_1^3 +r_2^3 + r_3^3=0$, therefore, we need $8-6a-3b = 0$, or equivalently $6a+3b = 8$.  But when $a$ and $b$ are integers, the left side is a multiple of 3, while $8$ is not a multiple of 3, so there is no solution; the cardinality of the set in question therefore is $0$.
