Exercise 1.1.6

Proof.

(a)

The polynomial $x^3+x-2=(x-1)(x^2+x+2)$ has a unique real root $1$, since the discriminant of $x^2+x+2$ is negative.

As $p∕3=1∕3,q∕2=-1$,

$$\begin{array}{llll}\hfill z_1& =\sqrt[3]{-\frac{q}{2}+\sqrt{{\left(\frac{q}{2}\right)}^2+{\left(\frac{p}{3}\right)}^3}}=\sqrt[3]{1+\sqrt{1+\frac{1}{27}}},& \hfill & \\ \hfill z_1& =\sqrt[3]{1+\frac{2}{3}\sqrt{\frac{7}{3}}},& \hfill & \\ \hfill z_2& =\sqrt[3]{1-\frac{2}{3}\sqrt{\frac{7}{3}}}.& \hfill & \end{array}$$

The roots of $x^3+x-2$ are

$$\begin{array}{llll}\hfill x_1& =z_1+z_2,& \hfill & \\ \hfill x_2& =\omega z_1+{\omega }^2{z}_2,& \hfill & \\ \hfill x_3& ={\omega }^2{z}_1+\omega z_2.& \hfill & \end{array}$$

As $z_1+z_2$ is real, it is the unique real root of $x^3+x-2$, so

$$1=\sqrt[3]{1+\frac{2}{3}\sqrt{\frac{7}{3}}}+\sqrt[3]{1-\frac{2}{3}\sqrt{\frac{7}{3}}}.$$

(b)

$$\begin{array}{llll}\hfill {\left(\frac{1}{2}+\frac{1}{2}\sqrt{\frac{7}{3}}\right)}^3& =\frac{1}{8}{\left(1+\sqrt{\frac{7}{3}}\right)}^3& \hfill & \\ \hfill & =\frac{1}{8}\left(1+3\sqrt{\frac{7}{3}}+3\frac{7}{3}+\frac{7}{3}\sqrt{\frac{7}{3}}\right)& \hfill & \\ \hfill & =1+\frac{1}{8}\left(3+\frac{7}{3}\right)\sqrt{\frac{7}{3}}& \hfill & \\ \hfill & =1+\frac{2}{3}\sqrt{\frac{7}{3}}.& \hfill & \end{array}$$

So $\sqrt[3]{1+\frac{2}{3}\sqrt{\frac{7}{3}}}=\frac{1}{2}+\frac{1}{2}\sqrt{\frac{7}{3}}$, and similarly $\sqrt[3]{1-\frac{2}{3}\sqrt{\frac{7}{3}}}=\frac{1}{2}-\frac{1}{2}\sqrt{\frac{7}{3}}$.

Consequently,

$$\sqrt[3]{1+\frac{2}{3}\sqrt{\frac{7}{3}}}+\sqrt[3]{1-\frac{2}{3}\sqrt{\frac{7}{3}}}=\frac{1}{2}+\frac{1}{2}\sqrt{\frac{7}{3}}+\frac{1}{2}-\frac{1}{2}\sqrt{\frac{7}{3}}=1.$$