Brian Bi
Exercise 15.2.1 A straightforward way to solve this problem uses Gaussian elimination. Observe that \begin{align*} (1 + \alpha + \alpha^2)1 &= 1 + \alpha + \alpha^2 \\ (1 + \alpha + \alpha^2)\alpha &= \alpha + \alpha^2 + \alpha^3 \\ &= \alpha + \alpha^2 + (3\alpha - 4) \\ &= -4 + 4\alpha + \alpha^2 \\ (1 + \alpha + \alpha^2)\alpha^2 &= (-4 + 4\alpha + \alpha^2)\alpha \\ &= -4\alpha + 4\alpha^2 + \alpha^3 \\ &= -4\alpha + 4\alpha^2 + (3\alpha - 4) \\ &= -4 - \alpha + 4\alpha^2 \end{align*} so that the equation $$(\alpha^2 + \alpha + 1)(a + b\alpha + c\alpha^2) = 1$$ can be written as the linear system $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 & -4 & -4 \\ 1 & 4 & -1 \\ 1 & 1 & 4 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ After row reduction, we find $$a = \frac{17}{49}, b = -\frac{5}{49}, c = -\frac{3}{49}$$.
Exercise 15.2.2 I don't know why Artin decided to alternate the signs, but this just makes the problem more annoying, so let's assume $$f(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_0$$. Observe that $$\alpha(\alpha^{n-1} + a_{n-1}\alpha^{n-2} + \ldots + a_1) = \alpha^n + a_{n-1}\alpha^{n-1} + \ldots + a_1\alpha = -a_0$$. Since $$f$$ is irreducible, it follows that $$a_0 \ne 0$$, so we may multiply both sides by $$-\frac{\alpha^{-1}}{a_0}$$ to obtain $$\alpha^{-1} = -\frac{1}{a_0}(\alpha^{n-1} + a_{n-1}\alpha^{n-2} + \ldots + a_1)$$.
Exercise 15.2.3 The minimal polynomial for $$\omega \sqrt[3]{2}$$ over $$\Q$$ is $$x^3 - 2$$. This is also the minimal polynomial for $$\sqrt[3]{2}$$. By Proposition 15.2.8, there is an isomorphism of field extensions $$\varphi : \Q(\omega \sqrt[3]{2}) \to \Q(\sqrt[3]{2})$$ that is the identity on $$\Q$$. If $$\sum_{i=1}^n x_i^2 = -1$$, then applying $$\varphi$$ to both sides yields $$\sum_{i=1}^n \varphi(x_i)^2 = -1$$. But each $$\varphi(x_i)$$ is real, so a contradiction has been reached.