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This involves factoring from Vieta's Formulas:

The first step is to break $a^3+b^3+c^3$ into $3r_1r_2r_3+(r_1+r_2+r_3)[r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1)$.

The confusing part might be trying to find $r_1^2+r_2^2+r_3^2$, yet we know that is equal to $(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1).$

This can be better written as $3r_1r_2r_3+(r_1+r_2+r_3)[(r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1)].$

Remember that $r_1, r_2$, and $r_3$ are the roots of the polynomial which is an expression.

Note that $r_1+r_2+r_3=\frac{-b}{a}$ , $r_1r_2+r_2r_3+r_3r_1=\frac{c}{a}$, and  $r_1r_2r_3=\frac{-d}{a}.$

Try to plug the values in, and be careful and don't forget the $x^2$ term.

 Let $a$, $b$, $c$ be the roots of $x^3 - 17x - 19$.  Find $a^3 + b^3 + c^3$.

$\begin{array}{|rcll|} \hline && x^3 - 17x - 19 \\ &=& x^3 + {\color{red}0}x^2 {\color{green}-17}x {\color{blue}-19} \\\\ \hline \mathbf{\text{vieta:}}\\ {\color{red}0} &=& -(a+b+c) \\ \mathbf{a+b+c} &=& \mathbf{0} \\\\ {\color{green}-17} &=& ab+ac+bc \\ \mathbf{ab+ac+bc} &=& \mathbf{-17} \\\\ {\color{blue}-19} &=& -abc \\ \mathbf{abc} &=& \mathbf{19} \\ \hline \end{array}$

 Formula:

$\begin{array}{|rcll|} \hline a^3+b^3+c^3 &=& 3\underbrace{abc}_{=19}+\underbrace{(a+b+c)}_{=0}\left(a^2+b^2+c^2-(ab+bc+ca)\right) \\ a^3+b^3+c^3 &=& 3\times 19 \\ \mathbf{a^3+b^3+c^3} &=& \mathbf{57} \\ \hline \end{array}$