Use a computer to find the three solutions of the equation${\mathbf{x}}^3\mathbf{-}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{0}$ . Find a way to show that the solutions can be writtenas .$\mathbf{2}\mathbf{cos}\mathbf{\left(}\frac{\pi }{9}\mathbf{\right)}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{cos}\mathbf{\left(}\frac{2\pi }{9}\mathbf{\right)}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{cos}\mathbf{\left(}\frac{4\pi }{9}\mathbf{\right)}$

For any complex numbers, let say,$a\text{and}b$ the definition of the complex power induces a formula as: $a^b=e^{blna}$, where.$a\ne e$

The given polynomial is$x^3-3x-1=0$, with roots .$2\cos \left(\frac{\pi }{9}\right),-2\cos \left(\frac{2\pi }{9}\right),\text{and\hspace{0.17em}}-2\cos \left(\frac{4\pi }{9}\right)$

Using computer, the three roots obtained are:

$\begin{array}{c}{x}_1=1.88\\ x_2=-0.347\\ x_3=-1.53\end{array}$

Let these roots are real part of the complex roots$z_1,z_2\text{and}{z}_3$given by:

$\begin{array}{l}{z}_1=x_1+i{y}_1=2\{\cos {\theta }_1+i\sin {\theta }_1\}\\ z_2=x_2+i{y}_2=2\{\cos {\theta }_2+i\sin {\theta }_2\}\\ z_3=x_3+i{y}_3=2\{\cos {\theta }_3+i\sin {\theta }_3\}\end{array}$

From the equation for $z_1$solve for the roots as:

$\begin{array}{c}{x}_1=2\cos {\theta }_1=1.88\\ {\theta }_1={\cos }^{-1}\left[\frac{1.88}{2}\right]=\pm \frac{\pi }{9}\\ Þx_1=2\cos \left(\frac{\pi }{9}\right)\end{array}$

From the equation for$z_2$solve for the roots as:

$\begin{array}{c}{x}_2=2\cos {\theta }_2=-0.347\\ {\theta }_2={\cos }^{-1}\left[\frac{0.347}{2}\right]=\pm \frac{4\pi }{9}\\ \Rightarrow x_2=-2\cos \left(\frac{4\pi }{9}\right)\end{array}$

From the equation for$z_3$solve for the roots as:

$\begin{array}{c}{x}_3=2\cos {\theta }_3=-1.53\\ {\theta }_3={\cos }^{-1}\left[\frac{1.53}{2}\right]=\pm \frac{2\pi }{9}\\ \Rightarrow x_3=-2\cos \left(\frac{2\pi }{9}\right)\end{array}$

Hence, the solutions can be written as:

$\begin{array}{c}{x}_1=1.88=2\cos \left(\frac{\pi }{9}\right)\\ x_2=-0.347=-2\cos \left(\frac{4\pi }{9}\right)\\ x_3=-1.53=-2\cos \left(\frac{2\pi }{9}\right)\end{array}$