The Roots Are Right There!

$x^3 + ax^2 + bx + c = 0$

According to Vieta's formula, if $a$ , $b$ , & $c$ are the roots of the equation, we can formulate these constraints for the coefficients:

$a + b + c = -a$ __(1)

$ab + bc + ca = b$ __(2)

$abc = -c$ __(3)

From the third constraint, it is clear that $ab = -1$ , and since these numbers are all integers, the values only vary between $1$ & $-1$ . In other words, if $a = 1$ , then $b = -1$ , or if $a = -1$ then $b = 1$ .

Either way, the sum of $a + b = 0$ , so within the second constraint, we can calculate for $a$ :

$ab + bc + ca = b$

$ab + c (b + a ) = ab = b$

Thus, $a = 1$ . Then $b = -1$ , and within the first constraint, $c = -a = -1$ .

As a result, the original equation is $x^3 + x^2 - x -1 = (x - 1)(x + 1)(x + 1) = 0$ .

Now for the quadratic equation, we can write as: $x^2 - x -1 = 0$

In order to evaluate $\alpha^2 + \beta^2$ , we can rewrite it as the following:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2(\alpha \beta)$

Again by using Vieta's formula, we know that $\alpha + \beta = \dfrac{-b}{a} = -(-1) = 1$ and $\alpha \beta = \dfrac{c}{a} = -1$ .

As a result, $\alpha^2 + \beta^2 = (1)^2 - 2(-1) = \boxed{3}$ .