Solve for all possible values of the real numbers x and y in the following equations.

${\left(x+iy\right)}^{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{1}$

It is given that ${\left(x+iy\right)}^3=-1$.

Every complex number can be represented as:

z=a+bi

Whereaandbare both real numbers, z is the complex number, and i is known as iota, which makes z a complex number.

State the given information and simplify it as:

$$\begin{gathered} {\left(x+iy\right)}^3=-1 \\ \left(x+iy\right){\left(x+iy\right)}^2=-1 \\ \left(x+iy\right)\left(x^2-y^2+2xyi\right)=-1 \\ {x}^3-x{y}^2+2x^{2}yi+x^{2}yi-y^{3}i-2x{y}^2=-1 \end{gathered}$$

Equate like terms from both sides:

$$\begin{gathered} x^3-3x{y}^2=-1 \\ 3x^{2}y-y^3=0 \end{gathered}$$

Simplify the second equation, and we obtain:

$$\begin{gathered} \left(3x^2-y^2\right)y=0 \\ y=0Or{y}^2=3x^2. \end{gathered}$$

Substitute$y=0$ in the first equation.

$$\begin{gathered} x^3=-1 \\ x=-1 \end{gathered}$$

Thus, the first answer is:

$$\begin{gathered} z_1=x_1+i{y}_1 \\ =-1 \\ =\left(-1,0\right) \end{gathered}$$

Use the second possibility of y, $y^2=3x^2$, and substitute it in the first equation and simplify:

$$\begin{gathered} x^3-3x\left(3x^2\right)=-1 \\ -8x^3=-1 \\ x=0.5 \end{gathered}$$

Substitute this back in the equation for y as:

$$\begin{gathered} y^2=3{\left(0.5\right)}^2 \\ {y}^2=\frac{3}{4} \\ y=\pm \frac{\sqrt{3}}{2} \end{gathered}$$

Thus, the answers are obtained:

$$\begin{gathered} z_2=x_2+i{y}_2=\frac{1}{2}+\frac{\sqrt{3}}{2}i \\ {z}_3=x_3+i{y}_3=\frac{1}{2}+\frac{\sqrt{3}}{2}i \end{gathered}$$

Hence, the final answers are obtained:

$$\begin{gathered} \left(x_1,y_1\right)=\left(-1,0\right) \\ \left(x_2,y_2\right)=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right) \\ \left(x_3,y_3\right)=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right) \end{gathered}$$