<s>[INST] Vieta's Formula - Hard Problem 
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 $(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}$.

How would I compute this? How to factor it?
 [/INST] We begin by observing that $x^2+6x+9 = (x+3)^2$ 
Denote $f(x) = (x^2+6x+9)^{50} - 4x +3 = (x+3)^{100}-4x+3$
$\{r_i\}_{1\leq i \leq 100}$ are roots of this polynomial so $f(r_i)= (r_i+3)^{100}-4r_i+3 = 0$ $\forall 1 \leq i \leq 100$
By Vieta's formula, the sum of roots is negative of coefficient of $x^{99}$ term in $f(x)$ which by Binomial Expansion turns out to be $-\binom{100}{1}3^1 = -300$
Now, we just need to make some manipulations, as follows:
$$(r_i+3)^{100} - 4r_i + 3 = 0 \implies (r_i+3)^{100} = 4r_i - 3$$
Therefore, 
$$\sum_{i=1}^{100} (r_i+3)^{100} = \sum_{i=1}^{100} 4r_i - 3 = 4 \left(\sum_{i=1}^{100} r_i\right) -300 = -1500$$
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