If $f(x) =ax^3+bx^2+cx+d$ is a cubic equation with roots $\alpha,\beta,\gamma.$ Is there a way to find $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha?$

Question

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 $\alpha^2\gamma+\gamma^2\beta+\beta^2\alpha = \gamma^2\beta+\beta^2\alpha+\alpha^2\gamma = \beta^2\alpha + \alpha^2\gamma+\gamma^2\beta.$

Answer 1

You know the elementary symmetric polynomials evaluated at the roots (by the Vieta relations): $$ \begin{aligned}s_1&=\alpha+\beta+\gamma=-b/a,\\ s_2&=\alpha\beta+\beta\gamma+\gamma\alpha=c/a,\\ s_3&=\alpha\beta\gamma=-d/a. \end{aligned} $$ The fundamental theorem of symmetric polynomials says that every symmetric polynomial can be written in terms of the elementary ones.

You are interested in finding $$ \begin{aligned} u&=\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha,&\text{or}\\ v&=\alpha^2\gamma+\gamma^2\beta+\beta^2\alpha. \end{aligned} $$ The problem, as explained by the others, is that you cannot tell which is which because $u$ and $v$ only follow cyclic symmetry.

However, the combinations $u+v$ and $uv$ are fully symmetric. A banal (but a bit tedious) calculation shows that $$ \begin{aligned} u+v&=s_1s_2-3s_3,\\ uv&=s_1^3s_3-6s_1s_2s_3+s_2^3+s_3^2. \end{aligned} $$ This means that we know the coefficients of the quadratic polynomial $$ p(x)=(x-u)(x-v)=x^2-[u+v]x+uv $$ that has $u$ and $v$ as its roots. All you need to do is plug in the known values of $s_1,s_2,s_3$ into the formulas above, and solve the quadratic $p(x)=0$. The roots are the two choices.

Answer 2

A function of two or more variables is said to be symmetric function if $f$ remains unaltered by an interchange of any two of it's variables.

If $\alpha, \beta, \gamma$ are roots of a cubic equation and the function $f$ of $\alpha,\beta,\gamma$ is invariant under the permutations of $\alpha, \beta, \gamma$, then we call $f$ , a symmetric function of roots.

Let us consider a function

$f(x, y, z) =x^2y+y^2x+z^2x+x^2z+y^2z+z^2y$

Then $f$ is a symmetric function of the roots $\alpha, \beta, \gamma$.

Here comes the nice formula

$\sum_{\text{sym}} \alpha^2\beta= \sum_{\text{sym}} \alpha \sum_{\text{sym}} \alpha\beta-3\alpha\beta\gamma$

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Now consider the function

$f(x, y, z) =x^2y+y^2z+z^2x$

Then $f$ is not a symmetric function of the roots $\alpha, \beta, \gamma$.

The values of $f$ changes along with the permutations of the roots.

Hence $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$ may have different values.

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$f(\alpha, \beta, \gamma) =\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$ have two different vaues for distinct values of $\alpha,\beta,\gamma$. One for all even permutations and other for all odd permutations.

For even permutations :

$f(\alpha, \beta, \gamma) =f(\beta, \gamma, \alpha) =f(\gamma, \alpha, \beta) $

For odd permutations :

$f(\alpha, \gamma, \beta) =f(\beta, \alpha, \gamma) =f(\gamma, \beta, \alpha) $

But still the sum depends on the choice of the order of the roots $\alpha, \beta, \gamma$.

Answer 3

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 $\alpha$, $\beta$, $\gamma$.