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Note that 

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

 

For each of \(i=1,2,\cdots,100\), since \(x = r_i\) is a root of \(f(x)\):

\(\begin{array}{rcl} f(r_i) &=& 0\\ (r_i+ 3)^{100} &=& 4r_i - 3 \end{array}\)

 

Now, using the equation above,

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

 

By Vieta's formula, we have:

\(r_1 + r_2 + \cdots + r_{100} = -\dfrac{\text{coefficient of }x^{99}\text{ in }f(x)}{\text{coefficient of }x^{100}\text{ in }f(x)}\)

 

Using the binomial theorem, we can expand \((x + 3)^{100}\).

\((x + 3)^{100} -4x+3 = x^{100} + 300x^{99}+\cdots\)

 

Therefore, 

\(r_1 + r_2 + \cdots + r_{100} = -\dfrac{300}1 = -300\)

 

Recall that \((r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}=4(r_1 + r_2 + \cdots + r_{100}) - 300\). Then:

\(\begin{array}{cl} &(r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}\\ =&4(-300) -300\\ =& -1500 \end{array}\)