Exercise 1.3.12

Proof.

(a)

Let $a,b\in \mathbb{Z}$ : $$\begin{array}{llll}\hfill {(a+\mathit{ib}\sqrt{11})}^3& =a^3+3i{a}^{2}b\sqrt{11}-3\times 11a{b}^2-i{b}^{3}11\sqrt{11}& \hfill & \\ \hfill & =(a^3-33a{b}^2)+i\sqrt{11}(3a^{2}b-11b^3),& \hfill & \end{array}$$

so, using the irrationality of $\sqrt{11}$,

$$4+i\sqrt{11}={(a+\mathit{ib}\sqrt{11})}^3⟺\{\begin{array}{ccc}\hfill 4\hfill & \hfill =\hfill & \hfill a^3-33a{b}^2,\hfill \\ \hfill 1\hfill & \hfill =\hfill & \hfill 3a^{2}b-11b^3.\hfill \end{array}$$

(b)

The first equation show that $a\mid 4$, and the second that $b\mid 1$, thus $b=\pm 1$.

As $b^3=b=\pm 1$, the second equation gives $3a^2-11=\pm 1$, thus $3a^2=10$ or $3a^2=12$. $3a^2=10$ is impossible since $3\nmid 10$. $3a^2=12$ gives $a=\pm 2$. But then $4=a^3-33a{b}^2=a(a^2-33)=\pm 2\times 29$, which is false.

The equation $4+i\sqrt{11}={(a+\mathit{ib}\sqrt{11})}^3$ has no solution.

(c)

We must find $p,q$ such that $-\frac{q}{2}=4,{\left(\frac{q}{2}\right)}^2+{\left(\frac{p}{3}\right)}^3=-11$.

$q=-8,p=-9$ is a solution.

An equation with solution $\sqrt[3]{4+\sqrt{11}i}+\sqrt[3]{4-\sqrt{11}i}$ is so

$$y^3-9y-8=0.$$