Vieta is WOW

Similar solution as @Aaryan Maheshwari's

Let the three roots of the cubic equation be $\alpha$ , $\alpha$ , and $\beta$ . Then by Vieta's formula :

$\begin{cases} 2\alpha + \beta = -p & ...(1) \\ \alpha^2 + 2\alpha \beta = 0 & ...(2) \\ \alpha^2 \beta = -q & ...(3) \end{cases}$

From $2 \alpha \times (1) - (2)$ : $\implies 3 \alpha^2 = - 2\alpha p \implies \alpha = - \dfrac 23 p \quad ...(2a)$

From $- \dfrac {27}2 \alpha \times (2)$ :

$\begin{aligned} - \frac {27}2 {\color{#3D99F6} \alpha^3} - 27 \color{#D61F06} \alpha^2 \beta & = 0 & \small \color{#3D99F6} \text{Note that }(2a): \ \alpha = - \frac 23 p \\ - \frac {27}2 \left({\color{#3D99F6} - \frac 8{27}p^3}\right) + 27 \color{#D61F06} q & = 0 & \small \color{#D61F06} \text{and }(3): \ \alpha^2 \beta = - q \\ 4p^3 + 27 q & = \boxed 0 \end{aligned}$

Let the roots of the equation be a, a, b. Then a(a+2b)=0. Since a can not be zero (because q is not zero), therefore a=-2b. 2a+b=-p and (a^2)b=-q. These imply 4(p^3)+27q=0

let the common root be 'a'

differentiating f(x)=x^3+px^2+q=0

f'(x)=3x^2+2px=0

roots of f'(x) are 0 and -2/3p

one of this roots will be normal point of inflection and other will be common root

as 0 cannot be a root hence a=-2/3p

as a is root of f(x) hence f(a)=0

$\frac{-8p^3}{27}$ + $\frac{4p^3}{9}$ +q=0

$\frac{(12-8)p^3}{27}$ +q=0

4p^3+27q=0

setting $x^3+px^2+q=(x-a)(x-b)^2$ leads to

$x^3+px^2+q=x^3-(2b+a)x^2+(b^2 +2ab)x-ab^2$

so that, by comparing the coefficient for each power of x:

$p=-(2b+a)$ , $0=b^2 +2ab$ , and $q=-ab^2$ .

substituting $b=-2a$ gives

$p=3a, q=-4a^3$ so that we can calculate $4p^3+27q=108a^3-108a^3=\boxed{0}$