Rational $3$-torsion points of an elliptic curve. Let $E$ be an elliptic curve defined over $\mathbb{Q}$ with equation $y^2=x^3+(ax+b)^2$, where $a,b \in \mathbb{Q}$. 
I have to prove that $0$ and $(0, \pm b)$ are the rational $3$-torsion points of $E$.
I know that if $Q$ is a $3$-torsion point then $[2]Q=-Q$ and that the $x$-coordinate of a $3$-torsion point satisfies $x(3x^3+4a^2x^2+12abx+12b^2)=0$ 
Now, clearly $0$ is a $3$-torsion point and if I choose $x=0$ and I plug it into the equation of $E$ I get $y^2=b^2$, thus $y= \pm b$. 
But now I don't know to continue and prove that there are no other rational $3$-torsion points (for example if $a,b \in \mathbb{Z}$ I would have used the rational root theorem, but I don't know in this case..)

Now $\Delta_E=16b(4a^3b^2-27b^3)$. 
We want to find the solution for $3x^3+4a^2x^2+12abx+12b^2=0$ and we can rewrite it as 
$3(x^3+a^2x^2+2abx+b^2)+a^2x^2+6abx+9b^2=3y^2+(ax+b)^2$ and this is equal to $0$ iff $y=0$ and $x=-\frac{3b}{a}$. 
Thus, if we plug those values to the equation of $E$ we get $0=-\frac{27b^3}{a^3}+4b^2 \Leftrightarrow 4a^3b^2-27b^3=0 \Leftrightarrow \Delta_E=0 $ but then we get a singular curve, hence it cannot be.
