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# help

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

Nov 27, 2019

#1
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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.

Nov 27, 2019
#2
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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:

$$\begin{array}{|rcll|} \hline a^3+b^3+c^3 &=& 3\underbrace{abc}_{=19}+\underbrace{(a+b+c)}_{=0}\left(a^2+b^2+c^2-(ab+bc+ca)\right) \\ a^3+b^3+c^3 &=& 3\times 19 \\ \mathbf{a^3+b^3+c^3} &=& \mathbf{57} \\ \hline \end{array}$$ Nov 28, 2019