Vieta kills it

I'm going to be using a different but more interesting (at least to me!) approach.

From Tom's explanation, we can note that the equation can be rewritten as $(x-a)^2(x+2a)$ . Now, examine the equation $y^2=x^3+qx+r$ . This is an elliptic curve. Note that the elliptic curve can be rewritten as $y^2=(x-a)^2(x+2a)$ .

However, it is clear that the elliptic curve has a singular point at $(a, 0)$ , making the curve singular. Recall that the definition of elliptic curves of the form $y^2=x^3+px+r$ traditionally excludes singular curves (and in a singular curve, $4q^3+27r^2=0$ ), giving us our answer.

Let the above cubic be expressed as $(x-a)^2(x-b) = (x^2-2ax+a^2)9x-b) = x^3 - (2a+b)x^2 + (a^2 + 2ab)x - a^2b = x^3 + qx+r$ , which yields:

$2a+b = 0;$

$a^2+2ab = q;$

$-a^2b = r$ .

If $b = -2a$ , then substituting this value into the remaining two equations gives:

$a^2 +2a(-2a) = q \Rightarrow \boxed{q = -3a^2}$ ;

$-a^2(-2a) = r \Rightarrow \boxed{r = 2a^3}$

which satisfy Choice D above: $4q^3 + 27r^2 = 4(-3a^2)^3 + 27(2a^3)^2 = -108a^6 + 108a^6 = 0.$

Let the roots of $x^3+qx+r =0$ be $(\alpha, \alpha, \beta)$ . By Vieta's formula , we have:

$\begin{cases} 2\alpha + \beta = 0 & \implies \blue{\beta = - 2\alpha} \\ q = \alpha^2 + 2\alpha \beta = \alpha + 2 \alpha \blue{(-2\alpha)} = - 3\alpha^2 & \implies q^3 = -27 \alpha^6 \\ r = -\alpha^2 \blue \beta = -\alpha^2 \blue{(-2\alpha)} = 2\alpha^3 & \implies r^2 = 4 \alpha^6 \end{cases}$

$\implies \boxed{4q^3 + 27r^2 = 0}$