Question regarding roots of a cubic polynomial

Question

If $\alpha$, $\beta$ and $\gamma$ are the roots of a cubic equation with

$$\alpha + \beta + \gamma = 1$$ $$\alpha^2 + \beta^2 + \gamma^2 = 2$$ $$\alpha^3 + \beta^3 + \gamma^3 = 3$$

Then find the value of

$$\alpha^4 + \beta^4 + \gamma^4$$

Answer 1

\begin{align} & \alpha \beta +\alpha \gamma +\beta \gamma =\frac{{{(\alpha +\beta +\gamma )}^{2}}-({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}})}{2}=-\frac{1}{2} \\ & \alpha \beta \gamma =\frac{({{\alpha }^{3}}+{{\beta }^{3}}+{{\gamma }^{3}})-(\alpha +\beta +\gamma )({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}-\alpha \beta -\alpha \gamma -\beta \gamma )}{3}=\frac{1}{6} \\ & {{\alpha }^{2}}{{\beta }^{2}}+{{\alpha }^{2}}{{\gamma }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}={{(\alpha \beta +\alpha \gamma +\beta \gamma )}^{2}}-2(\alpha \beta \gamma )(\alpha +\beta +\gamma )=-\frac{1}{12} \\ & {{\alpha }^{4}}+{{\beta }^{4}}+{{\gamma }^{4}}={{({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}})}^{2}}-2({{\alpha }^{2}}{{\beta }^{2}}+{{\alpha }^{2}}{{\gamma }^{2}}+{{\beta }^{2}}{{\gamma }^{2}})=4+\frac{1}{6}=\frac{25}{6} \\ \end{align}

Answer 2

I will use the notation of this question of mine. The $\text{EXP}$ map gives: $$ \exp\left(-\sum_{m\geq 1}\frac{p_m}{m}x^m\right) = \sum_{r=0}^{3}(-1)^r e_r x^r \tag{1}$$ hence a polynomial having $\alpha,\beta,\gamma$ as roots is: $$ p(x)=x^3-x^2-\frac{x}{2}-\frac{1}{6} \tag{2} $$ and the $\text{LOG}$ map gives:

$$ p_4 = -4 [x^4]\log\left(1 - x - x^2/2 - x^3/6\right)=\color{red}{\frac{25}{6}}.\tag{3}$$

Once we have $(2)$, we may also go this way: any root of $p(x)$ fulfils $x^4=x^3+\frac{x^2}{2}+\frac{x}{6}$, hence: $$ p_4 = p_3+\frac{p_2}{2}+\frac{p_1}{6} = 3+1+\frac{1}{6}.\tag{4}$$