The polynomial $f(x)=x^3-3x^2-4x+4$ has three real roots $r_1$, $r_2$, and $r_3$. Let $g(x)=x^3+ax^2+bx+c$ be the polynomial which has roots $s_1$, $s_2$, and $s_3$, where
\begin{align*}
s_1 &= r_1+r_2z+r_3z^2, \\
s_2 &= r_1z+r_2z^2+r_3, \\
s_3 &= r_1z^2+r_2+r_3z,
\end{align*}and $z=\dfrac{-1+i\sqrt3}2$. Find the real part of the sum of the coefficients of $g(x)$.
I know the sum of the coefficients is $g(1)$, $g(x)=(x-s_1)(x-s_2)(x-s_3)$, and $z^3=1$. This means $s_1z=s_2$, and $s_2z=s_3$. Since $s_1^3=s_2^3=s_3^3$, I have $g(x)=x^3-s_1^3$. Since the answer is $g(1)$, I need to calculate $$1-s_1^3.$$ I expanded $s_1^3$ to get $$s_1^3=r_1^3+r_1^2r_2z+3r_1^2r_3z+3r_1r_2^2z^2+6r_1r_2r_3+3r_1r_3^2z+r_2^3+3r_2^2r_3z+3r_2^2r_3z+3r_2r_3^2z^2+r_3^3.$$ I'm pretty sure using Vieta's can finish this, but I'm not sure where else to apply Vieta's other than $r_1r_2r_3$. I also tried substituting $z^2=-z-1$, but it didn't do much. I also tried using $(r_1+r_2+r_3)^2$, but this also failed. Could someone give me some guidance?
Thanks in advance!
Best Answer
Well, the real part of $1−s_1^3$ is to be calculated, thus we need to calculate $\frac{ 1−s_1^3 + \overline{1−s_1^3}}{2}$ where $\overline{z}$ is the complex conjugate of z.
$\frac{ 1−s_1^3 + \overline{1−s_1^3}}{2}$ $\implies$$\frac{ 2−(s_1^3 + \overline{s_1^3})}{2}$ $\implies$ $\frac{ 2−(s_1+ \overline{s_1})(s_1^2+\overline{s_1}^2-s_1\overline{s_1})}{2}$
Also,
$s_1= r_1+r_2z+r_3z^2$
$\overline{s_1}=r_1+r_3z+r_2z^2$ (as $z^3=1$)
On simplifying $ (s_1+ \overline{s_1})(s_1^2+\overline{s_1}^2-s_1\overline{s_1})$,we obain an expression $2\displaystyle\sum_{i=1}^{3} r_i^3-3\displaystyle\sum_{1\leq i , j\leq 3,(i≠j) } r_i r_j^2+12\displaystyle\prod_{i=1}^{3} r_i$ (where $r_i$ are roots of f(x)).
Can you proceed further from here?