Sum of Cubic Roots (Part 2)

Let $f(x) = x^3+x+1$ . We note that $f'(x) = 3x^2 + 1 > 0$ for all $x$ , implying that it is an increasing function. Therefore, $f(x) = 0$ has only one real solution. Let the real root be $x_1$ , then $x_2$ and $x_3$ are complex. Since the coefficients of $f(x)$ are real, $x_2$ and $x_3$ are conjugates or $x_3=\overline {x_2}$ . We note that $f(0)=1$ and $f(-1) = -1$ , implying the real root $-1 < x_1 < 0$ . Let $x_1 = - \alpha$ , where $\alpha = |x_1|$ is positive and real.

By Vieta's formula , we have:

$\begin{aligned} x_1 + x_2 + x_3 & = 0 \\ x_1 + x_2 + \overline{x_2} & = 0 + 0i & \small \color{#3D99F6} \text{Equating the real and imaginary parts} \\ - \alpha + \frac \alpha 2 + \beta i + \frac \alpha 2 - \beta i & = 0 + 0i & \small \color{#3D99F6} \text{where }\beta \text{ is real,} \end{aligned}$

Implying that $x_2 = \dfrac \alpha 2 + \beta i$ and $x_2 = \dfrac \alpha 2 - \beta i$ .

Again, by Vieta's formula, we have:

$\begin{aligned} x_1x_2 x_3 & = - 1 \\ -\alpha x_2 \overline{x_2} & = - 1 \\ |x_2|^2 & = \frac 1\alpha \\ \implies |x_2| = |x_3| & = \frac 1{\sqrt \alpha} \\ \implies |x_1| + |x_2| + |x_3| & = \alpha + \frac 2{\sqrt \alpha} & \small \color{#3D99F6} \text{By numerical method }\alpha \approx 0.682327804 \\ & \approx 0.682327804 + \frac 2{\sqrt{0.682327804}} \\ & \approx \boxed{3.1035} \end{aligned}$