Let , and let be the roots of . Compute ​.

Let
\(f(x)=(x^2+6x+9)^{50}-4x+3\), and let \(r_1,r_2,\ldots,r_{100}\) be the roots of \(f(x)\).
Compute \((r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}\).

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x^2+6x+9)^{50}-4x+3} \\ &=& \left((x+3)^2\right)^{50}-4x+3 \\ \mathbf{f(x)} &=& \mathbf{\left( x+3 \right)^{100}-4x+3} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline & f(r_1) =0 &=& \left( r_1+3 \right)^{100}-4r_1+3 \\ & f(r_2) =0 &=& \left( r_2+3 \right)^{100}-4r_2+3 \\ & \ldots \\ & f(r_{100}) =0 &=& \left( r_{100}+3 \right)^{100}-4r_{100}+3 \\ & \hline \\ \text{sum} & 0 &=& (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} \\ & && -4(r_1+r_2+\ldots + r_{100}) + 3\cdot 100 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ \hline \end{array} \)

\(\mathbf{\text{vieta:}}\)

For any polynomial equation
\(0=x^n+a_{n-1}·x^{n-1}+...+a_2·x^2+a_1·x^1+a_0 \)
with the solutions \(r_1\dots r_n\), the relatively simple formulas for \(a_0\) and \(a_{n-1}\) are:

   \(a_0=(-1)^n \prod\limits_{k=1}^{n} r_k \\ a_{n-1}= -\sum \limits_{k=1}^{n} r_k\)

\(\begin{array}{|rcll|} \hline && \left( x+3 \right)^{100} \\ \\ &=& x^{100} + \binom{100}{1}x^{99}\cdot 3 + \ldots \\ \\ &=& x^{100} + \underbrace{300}_{ \underbrace{=a_{n-1}}_{= -(r_1+r_2+\ldots + r_{100})}} x^{99} + \ldots \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ && \boxed{r_1+r_2+\ldots + r_{100} = -300} \\ (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(-300) - 300 \\ \mathbf{(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}} &=& \mathbf{-1500} \\ \hline \end{array} \)

Thank you i understand but for Vieta's I learned that (-1)^k=(an-k/an) is this the same or sound familiar