Question

$x^{3} - ax + b = 0\sum_{\gamma}^{\alpha}\mspace{2mu}\ \alpha + \beta + \gamma = 0$

$$\begin{matrix} \alpha\beta + \beta\gamma + \gamma\alpha = - a \\ \alpha\beta\gamma = - b \end{matrix}$$

Equation will have 2 positive and 1 negative root.

$$\begin{matrix} & \text{~Let~}\ \begin{matrix} \gamma < 0 & \ < \alpha \leq \beta\text{~and~}|\alpha| \leq |\beta| \leq |\gamma| \\ b - a\alpha & \ = - \alpha\beta\gamma + \alpha(\alpha\beta + \beta\gamma + \gamma\alpha) \\ & \ = \alpha^{2}\beta + \alpha^{2}\gamma \\ & \ = - \alpha^{3} < 0 \\ \frac{b}{a} & \ < \alpha \\ 3b - 2a\alpha & \ = - 3\alpha\beta\gamma + 2\alpha(\alpha\beta + \beta\gamma + \gamma\alpha) \\ & \ = - \alpha\beta\gamma + 2\left( \alpha^{2}\beta + \alpha^{2}\gamma \right) \\ & \ = \alpha\left( - 2(\beta + \gamma)^{2} - \beta\gamma \right) \\ & \ = - \alpha\left( 2\beta^{2} + 2\gamma^{2} + 5\beta\gamma \right) \\ & \ = - \alpha(2\beta + \gamma)(\beta + 2\gamma) \\ & \ = - \alpha(\beta - \alpha)(\gamma - \alpha) \leq 0 \\ \Rightarrow \ \frac{3b}{2a} & \ \leq \alpha \end{matrix} \end{matrix}$$