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?
$x_1^3$ + $x_3^3$ + $x_2^3$
On
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 $S_1=-\frac ba$ [1], 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.
[1] we could even have deduced this from the linear equation $aS_1+b=0$
Use that $$(a+b+c)^3=a^3+b^3+c^3+3ab(a+b)+3ac(a+c)+3bc(b+c)+6abc$$ $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$