Inspired by Issac Newton

Finding $f(x)$ :

If $f(x)=(x+1)(x-1)(x-3)+a(x-1)(x-3)+b(x-3)+c$ , then: $\begin{cases}8a\!\!\!\!&-4b\!\!\!\!&+c\!\!\!\!&=-9\\&-2b\!\!\!\!&+c\!\!\!\!&=-7\\&&\ \ \ c\!\!\!\!&=19\end{cases}\ \Longleftrightarrow\ \begin{cases}a=3\\b=13\\c=19\end{cases}$ $\Longrightarrow \ f(x)=((x+1+a)(x-1)+b)(x-3)+c=\boxed{x^3-8}$

Finding the value of the expression:

Clearly, $\alpha^3=\beta^3=\xi^3=8$ and $\alpha^{3k}+\beta^{3k}+\xi^{3k}=8^k+8^k+8^k=3·8^k$ .

The topmost expression is $\sum_{k=1}^{20}{3·8^k}=3·\sum_{k=1}^{20}{8^k}=3·8·\dfrac{8^{20}-1}{8-1}=\frac{24}{7}(8^{20}-1)=\frac{24}{7}(2^{60}-1)$ .

Modulo 17, as $2^4=16\equiv -1$ : the answer is $\frac{24}{7}(2^{60}-1)\equiv \frac{7}{7}\left((2^4)^{15}-1\right)\equiv(-1)^{15}-1=-2\equiv\boxed{15}$ .

Let $f\left( x \right) ={ x }^{ 3 }+b{ x }^{ 2 }+cx+d$

$\therefore \begin{cases} {b+c+d=-8} \\ {b-c+d=-8 } \\ {27+8b+3c+d=19} \end{cases}$

$\Rightarrow b=c=0,d=-8$

$\Rightarrow f\left( x \right) ={ x }^{ 3 }-8$

Now roots of $f\left( x \right)$ are $2,2\omega ,{ 2\omega }^{ 2 }$ where $\omega$ is cube root of unity.

$\therefore \sum _{ k=1 }^{ 20 }{ \left( { \alpha }^{ 3k }+{ \beta }^{ 3k }+{ \xi }^{ 3k } \right) } =\sum _{ k=1 }^{ 20 }{ \left( 2^{ 3k }+{ (2\omega ) }^{ 3k }+{ ({ 2\omega }^{ 2 }) }^{ 3k } \right) }$

$=\sum _{ k=1 }^{ 20 }{ 2^{ 3k }\left( 1+{ \omega }^{ 3k }+{ \omega }^{ 6k } \right) } =\sum _{ k=1 }^{ 20 }{ 2^{ 3k }\left( 3 \right) } \because { \omega }^{ 3 }=1$

$\sum _{ k=1 }^{ 20 }{ 2^{ 3k }\left( 3 \right) } \quad$ mod $17$ $=15$

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