Well, I am trying to solve the following system of equations for $\left(\text{a},\text{b},\text{c}\right)$:
$$ \begin{align*}\begin{cases} \text{a}+\text{b}+\text{c}&=\alpha_1\\ \\ \text{a}^2+\text{b}^2+\text{c}^2&=\alpha_2\\ \\ \text{a}^4+\text{b}^4+\text{c}^4&=\alpha_3 \end{cases}\end{align*} \tag1 $$
My work, I did the following working out:
If we square the first equation we get:
$$\left(\underbrace{\text{a}+\text{b}+\text{c}}_{=\space\alpha_1}\right)^2=\underbrace{\text{a}^2+\text{b}^2+\text{c}^2}_{=\space\alpha_2}+2\left(\text{a}\left(\text{b}+\text{c}\right)+\text{b}\text{c}\right)\tag2$$
Squaring the second equation gives:
$$\left(\underbrace{\text{a}^2+\text{b}^2+\text{c}^2}_{=\space\alpha_2}\right)^2=\underbrace{\text{a}^4+\text{b}^4+\text{c}^4}_{=\space\alpha_3}+2\left(\text{a}^2\left(\text{b}^2+\text{c}^2\right)+\text{b}^2\cdot\text{c}^2\right)\tag3$$
This leads to:
$$\alpha_1^2=\alpha_2+2\left(\text{a}\left(\text{b}+\text{c}\right)+\text{b}\text{c}\right)\space\Longleftrightarrow\space2\left(\text{a}\left(\text{b}+\text{c}\right)+\text{b}\text{c}\right)=\alpha_1^2-\alpha_2\tag4$$ $$\alpha_2^2=\alpha_3+2\left(\text{a}^2\left(\text{b}^2+\text{c}^2\right)+\text{b}^2\cdot\text{c}^2\right)\space\Longleftrightarrow\space2\left(\text{a}^2\left(\text{b}^2+\text{c}^2\right)+\text{b}^2\cdot\text{c}^2\right)=\alpha_2^2-\alpha_3\tag5$$
Squaring equation $(4)$, gives:
$$\left(2\left(\text{a}\left(\text{b}+\text{c}\right)+\text{b}\text{c}\right)\right)^2=2\cdot\underbrace{2\left(\text{a}^2\left(\text{b}^2+\text{c}^2\right)+\text{b}^2\cdot\text{c}^2\right)}_{=\space\alpha_2^2-\alpha_3}+8\text{abc}\left(\underbrace{\text{a}+\text{b}+\text{c}}_{=\space\alpha_1}\right)\tag6$$
So, we can see that:
$$\left(\alpha_1^2-\alpha_2\right)^2=2\left(\alpha_2^2-\alpha_3\right)+8\text{abc}\cdot\alpha_1\space\Longleftrightarrow\space\text{abc}=\frac{\left(\alpha_1^2-\alpha_2\right)^2-2\left(\alpha_2^2-\alpha_3\right)}{8\alpha_1}\tag7$$
And from this point on I am stuck.
