Solution: Dividing $x^{3}+3 x+3$ by $x^{2}-2 x+7$ gives the quotient $x+2$ and remainder $-11$. Thus $u^{3}+3 u+3=(u+2)\left(u^{2}-2 u+7\right)-11$, and so $\left(7-2 u+u^{2}\right)^{-1}=(2+u) / 11=(2 / 11)+(1 / 11) u .$

(a) Show that $\mathbf{Q}(\sqrt{2}+i)=\mathbf{Q}(\sqrt{2}, i)$.
Solution: Let $u=\sqrt{2}+i$. Since $(\sqrt{2}+i)(\sqrt{2}-i)=2-i^{2}=3$, we have $\sqrt{2}-i=3(\sqrt{2}+i)^{-1} \in \mathbf{Q}(u)$, and it follows easily that $\sqrt{2} \in \mathbf{Q}(u)$ and
$i \in \mathbf{Q}(u)$, so $\mathbf{Q}(\sqrt{2}, i) \subseteq \mathbf{Q}(u) .$ The reverse inclusion is obvious.
(b) Find the minimal polynomial of $\sqrt{2}+i$ over $\mathbf{Q}$. Solution: We have $\mathbf{Q} \subseteq \mathbf{Q}(\sqrt{2}) \subseteq \mathbf{Q}(\sqrt{2}, i)$. Thus $[\mathbf{Q}(\sqrt{2}): \mathbf{Q}]=2$ since $\sqrt{2}$
is a root of a polynomial of degree 2 but is not in $\mathbf{Q} .$ We have $[\mathbf{Q}(\sqrt{2}, i)$ :
$\mathbf{Q}(\sqrt{2})]=2$ since $i$ is a root of a polynomial of degree 2 over $\mathbf{Q}(\sqrt{2})$ but is not in $\mathbf{Q}(\sqrt{2})$. Thus $[\mathbf{Q}(\sqrt{2}+i): \mathbf{Q}]=4$, and so the minimal polynomial for $\sqrt{2}+i$ must have degree $4 .$

Since $u=\sqrt{2}+i$, we have $u-i=\sqrt{2}, u^{2}-2 i u+i^{2}=2$, and $u^{2}-3=2 i u$. Squaring again and combining terms gives $u^{4}-2 u^{2}+9=0$. Thus the minimal polynomial for $\sqrt{2}+i$ is $x^{4}-2 x^{2}+9$

Solution: Let $x=1+\sqrt[3]{2} .$ Then $x-1=\sqrt[3]{2}$, and so $(x-1)^{3}=2$, which yields $x^{3}-3 x^{2}+3 x-1=2$, and therefore $x^{3}-3 x^{2}+3 x-3=0$. Eisenstein’s criterion (with $p=3$ ) shows that $x^{3}-3 x^{2}+3 x-3$ is irreducible over $\mathbf{Q}$, so this is the required minimal polynomial.