Determine the real numbers $a$, $b$, $c$ such that $1$, $\frac1{1+\omega}$ and $\frac1{1+\omega^*}$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$ I am stuck on this question:


Let $1$, $\omega$ and $\omega^*$ be the cube root of unity.
a. Show that $\dfrac1{1+\omega}=-\omega$ and $\dfrac1{1+\omega^*}=-\omega^*$.
b. Determine the real numbers $a$, $b$, $c$ such that $1$, $\dfrac1{1+\omega}$ and $\dfrac1{1+\omega^*}$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$.
c. Hence, find $p(\omega)$ and $p(\omega^*)$.

So I was able to do part a by finding the roots in Cartesian forms, but I am not sure how to approach part b.

We use the familiar fact that
$$
(x - \alpha) (x - \beta) = x^{2} - (\alpha + \beta) x + \alpha \beta.
$$
For part a, note that $1, \omega, \omega^{*}$ are the three distinct roots of
$$
x^{3} - 1 = (x -1) (x^{2} + x + 1).
$$
Therefore $\omega, \omega^{*}$ are the roots of $x^{2} + x + 1$. Hence their product equals the constant coefficient $1$, so that
$$
\omega^{*} = \omega^{-1},
$$
and their sum equals $-1$, that is, the negative of the coefficient of $x$, so that
$$
\omega + \omega^{*} = -1,
$$
and your formulas follow.
For b, you may note that because of the above formulas,
$$
\frac{1}{1 + \omega} = - \omega,
\quad\text{and}\quad
\frac{1}{1 + \omega^{*}} = - \omega^{*}
$$
are the roots of $x^{2} - x + 1$, and so 
$$
1, - \omega, - \omega^{*}
$$
will be the roots of
$$
(x - 1) (x^{2} - x + 1)
=
x^{3} - 2 x^{2} + 2 x - 1.
$$
