The sum of the cubes of all the roots of the equation x ^ 4 - 3 x ^...

{x}^{4} - 3{x}^{3} - 2{x}^{2} + 3x + 1 = 0$</p><p>Dividing by${x}^{2}$as$x \ne 0$</p><p>${x}^{2} - 3x - 2 + \dfrac{3}{x} + \dfrac{1}{{x}^{2}} = 0$</p><p>${x}^{2} + \dfrac{1}{{x}^{2}} - 2 + 2 - 3\left(x - \dfrac{1}{x}\right) - 2 = 0$</p><p>${\left(x - \dfrac{1}{x}\right)}^{2} - 3\left(x - \dfrac{1}{x}\right) = 0$</p><p>$\begin{cases}x - \dfrac{1}{x} = 0, & x - \dfrac{1}{x} = 3\end{cases}$</p><p>$\begin{cases}{x}^{2} - 1 = 0, & {x}^{2} - 3x - 1 = 0\end{cases}$</p><p>$x = \pm 1$</p><p>For${x}^{2} - 3x - 1 = 0$let the roots are$\gamma \text{ and } \delta$</p><p>$\begin{cases}\alpha = 1, \beta = - 1, \gamma + \delta = 3, & \gamma\delta = - 1\end{cases}$</p><p>${\alpha}^{3} + {\beta}^{3} + {\gamma}^{3} + {\delta}^{3}$</p><p>$= 1 - 1 + \left(\gamma + \delta\right)\left({\left(\gamma + \delta\right)}^{2} - 3\gamma\delta\right)$</p><p>$= 3\left(9 - 3\left(- 1\right)\right)$= 36$