# Using Vieta's formula to find the sum of the roots for a given cubic equation.

Vieta's formula states that, if a cubic equation has three different roots, the following is true:

$$\begin{eqnarray*} x_1 + x_2 + x_3 &=& -b/a\\ x_1x_2 + x_1x_3 + x_2x_3 &=& c/a \\ x_1x_2x_3 &=& -d/a \end{eqnarray*}$$

Then, how is the following calculated?

$$x_1^3$$ + $$x_3^3$$ + $$x_2^3$$

• This doesn't depend on the roots being distinct, but it does require $a\neq 0$ so that it is a genuine cubic and not a quadratic or lower. – Mark Bennet Oct 13 '18 at 14:41

Use that $$(a+b+c)^3=a^3+b^3+c^3+3ab(a+b)+3ac(a+c)+3bc(b+c)+6abc$$ $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$

• When using this, how would I separate the products from the sums so that I can use Vieta's formulas? – Stefan Ivanovski Oct 13 '18 at 14:14
• For example $$(x_1+x_2+x_3)^3=(-\frac{b}{a})^3$$ and so on – Dr. Sonnhard Graubner Oct 13 '18 at 14:15
• I meant, the case for: 3ab(a+b)+3ac(a+c)+3bc(b+c). How can I factor this so that it becomes in the form of one of the Vieta's formulas? (a+b+c | ab+bc+ac | abc) – Stefan Ivanovski Oct 13 '18 at 14:18
• $$3a^2(b+c)+3b^2(a+c)+3c^2(a+b)=3a^2(a+b+c-a)+3b^2(a+b+c-b)+3c^2(a+b+c-c)$$ – Dr. Sonnhard Graubner Oct 13 '18 at 14:29

If $$\alpha, \beta, \gamma$$ are the roots of $$p(x)=ax^3+bx^2+cx+d=0$$ then $$p(\alpha)=p(\beta)=p(\gamma)=p(\alpha)+p(\beta)+p(\gamma)=0$$

Now set $$S_n=\alpha^n+\beta^n+\gamma^n$$ and the last equation tells us that $$aS_3+bS_2+cS_1+3d=0$$

The sum of squares $$S_2$$ is easy to find from the symmetric polynomials, and $$S_1$$ is known. This observation avoids having to remember complicated formulae for the sums of powers. (We might put $$S_0=3$$)

Indeed for higher powers we can observe

$$\alpha^r p(\alpha)+\beta^r p(\beta)+\gamma^rp(\gamma)=0$$ which gives $$aS_{r+3}+bS_{r+2}+cS_{r+1}+dS_r=0$$ which enables us to compute the sums of powers successively.

Moreover, we can compute the formula for $$S_2$$ using a quadratic, as if only two roots were involved, so that $$aS_2+bS_1+2c=0$$ with $$S_1=-\frac ba$$ , so that $$S_2=\frac {b^2}{a^2}-2\left(\frac ca\right)$$ and this remains true for any larger number of roots. Similarly the formula for three roots which we can derive from the cubic applies for four or more roots too.

 we could even have deduced this from the linear equation $$aS_1+b=0$$