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 \((r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}\).
+9 \(f(x)=((x+3)^2)^{50}-4x+3=(x+3)^{100}-4x+3\)
\((r_i+3)^{100}-4r_i+3=0 (i=1, 2, ...,100)\)
\((r_i+3)^{100}=4r_i-3\)
Therefore, \((r_1+3)^{100}+(r_2+3)^{100}+...+(r_{100}+3)^{100}=(4r_1-3)+(4r_2-3)+...+(4r_{100}-3)\)
\( =4(r_1+r_2+...+r_{100})-300\)
\( =4*(-300)-300=-1500\)
(By Vieta's formulas, \(-(r_1+r_2+...+r_{100})\) equals the coefficient of \(x^{99}\) of \(f(x)\), which is \(nCr(100, 1)*3^1=300\) by the binomial theorem.
Therefore, \(r_1+r_2+...+r_{100}=-300\).)
