If $f(x) =ax^3+bx^2+cx+d$ is a cubic equation with roots $\alpha,\beta,\gamma.$ Is there a way to find $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha?$ Suppose $f(x) = ax^3 + bx^2 + cx + d$ is a cubic equation with roots $\alpha, \beta, \gamma.$ Then we have:
$\alpha + \beta + \gamma= -\frac{b}{a}\quad (1)$
$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\quad (2)$
$\alpha\beta\gamma = -\frac{d}{a}\quad (3)$
We can find $\alpha^2\beta + \beta^2\gamma + \gamma^2\alpha + \alpha^2\gamma + \gamma^2\beta + \beta^2\alpha$ in terms of $a,b,c,d$ with the formula:
$$ \alpha^2\beta + \beta^2\gamma + \gamma^2\alpha + \alpha^2\gamma + \gamma^2\beta + \beta^2\alpha = (\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma) - 3\alpha\beta\gamma $$
$$=\left(\frac{-b}{a}\right) \left(\frac{c}{a}\right) - 3\left(-\frac{d}{a}\right).$$
But I was wondering if there was some way to find $ \alpha^2\beta + \beta^2\gamma + \gamma^2\alpha\ $ and therefore also $\ \alpha^2\gamma + \gamma^2\beta + \beta^2\alpha\ $ in terms of $a,b,c,d,\ $ with some algebraic manipulation, i.e. without finding the roots with a cubic formula?
Notice that there are two possible values of $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha,$ namely $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha = \beta^2\gamma+\gamma^2\alpha+\alpha^2\beta = \gamma^2\alpha+\alpha^2\beta+\beta^2\gamma$ and $\alpha^2\gamma+\gamma^2\beta+\beta^2\alpha = \gamma^2\beta+\beta^2\alpha+\alpha^2\gamma = \beta^2\alpha + \alpha^2\gamma+\gamma^2\beta.$

Quantity is not-symmetric. This results in rather complex formula:
$$\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha=t_{1}\,t_{3}^2-{{b\,t_{3}^2}\over{3\,a}}+t_{2}^2\,t_{3}-{{2\,b\,
 t_{2}\,t_{3}}\over{3\,a}}-{{2\,b\,t_{1}\,t_{3}}\over{3\,a}}+\\{{b^2\,
 t_{3}}\over{3\,a^2}}-{{b\,t_{2}^2}\over{3\,a}}+t_{1}^2\,t_{2}-{{2\,b
 \,t_{1}\,t_{2}}\over{3\,a}}+{{b^2\,t_{2}}\over{3\,a^2}}-{{b\,t_{1}^2
 }\over{3\,a}}+{{b^2\,t_{1}}\over{3\,a^2}}-{{b^3}\over{9\,a^3}}$$
$$t_i=\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4}-\frac{p^3}{27}}}\,e^{k_i\frac{2i\pi}{3}}+\frac{p}{3\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4}-\frac{p^3}{27}}}}e^{-k_i\frac{2i\pi}{3}}, i\in\{1,2,3\}$$
$$p={{b^2}\over{3\,a^2}}-{{c}\over{a}}, q={{b\,c
 }\over{3\,a^2}}-{{d}\over{a}}-{{2\,b^3}\over{27\,a^3}}$$
$\{k_1,k_2,k_3\}=\{0,1,2\}$ determines order of roots taken as $\alpha$, $\beta$, $\gamma$.
