+4622 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.
+26393 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:
\(\begin{array}{|rcll|} \hline a^3+b^3+c^3 &=& 3\underbrace{abc}_{=19}+\underbrace{(a+b+c)}_{=0}\left(a^2+b^2+c^2-(ab+bc+ca)\right) \\ a^3+b^3+c^3 &=& 3\times 19 \\ \mathbf{a^3+b^3+c^3} &=& \mathbf{57} \\ \hline \end{array}\)
