Find the value of $α$

Let $f(x$ be the cubic in question. Then $f(x)=x^3+2x^2+3x+4=(x-a)(x-b)(x-c)$ . In particular, $f(-\alpha)=-(a+\alpha)(b+\alpha)(c+\alpha)=-\alpha^2$ (from the information in the question).

Combining all this, we find $\alpha^3-3\alpha^2+3\alpha-4=0$ . This is just $(\alpha-1)^3-3=0$ , which has real root $\alpha=1+3^{1/3}$ giving the answer $\boxed4$ .

$x^3 + 2x^2 + 3x + 4 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + bc + ac)x - abc \implies$

$a + b + c = -2$

$ab + bc + ac = 3$

$abc = -4$

and

$\alpha^2 = (a + \alpha)(b + \alpha)(c + \alpha) = \alpha^3 - (a + b + c)\alpha^2 + (ab + bc + ab)\alpha + abc \implies$

$\alpha^2 = \alpha^3 - 2\alpha^2 + 3\alpha - 4 \implies \alpha^3 - 3\alpha^2 + 3\alpha - 4 = 0$

Let $\alpha = u + 1 \implies u^3 + 3u^2 + 3u + 1 - 3(u^2 + 2u + 1) + 3(u + 1) - 4 = 0 \implies u^3 - 3 = 0$

$\implies u = 3^{\frac{1}{3}}$ for real $u \implies \alpha = 1 + 3^{\frac{1}{3}} = p + q^{\frac{1}{q}} \implies p + q = \boxed{4}.$

Simplifying $(a + \alpha)(b+\alpha)(c+\alpha) = \alpha^2$ gives $\alpha^3 + \alpha^2 (a + b + c - 1) + \alpha(ab + ac + bc) + abc = 0$

By Vieta's formula, $a + b + c = -2, ab + ac + bc = 3, abc = -4$ . Substitute these values into the equation above:

$\alpha^3 -3 \alpha^2 + 3\alpha - 4 = 0 \quad\Leftrightarrow \quad (\alpha - 1)^3 = 3 \quad\Leftrightarrow \quad \alpha = 1 + 3^{1/3}$

Then $p = 1, q = 3$ . The answer is $p+q=\boxed4$ .