Let $z=18+26i$ where $z_0=x_0+iy_0, \ x_0, y_0\in \Bbb R$ is the cube root of $z$ having least positive argument. Find the value of $x_0y_0(x_0+y_0)$. Let $z=18+26i$ where $z_0=x_0+iy_0, \quad x_0,y_0\in R$ is the cube root of $z$ having least positive argument.Find the value of $x_0y_0(x_0+y_0)$.

Let $\sqrt[3]{18+26i}=x_0+iy_0$
$18+26i=x_0^3+3ix_0^2y_0-3x_0y_0^2-iy_0^3$
$18=x_0^3-3x_0y_0^2,26=-y_0^3+3x_0^2y_0$.
I am stuck here. What should i do to find $x_0y_0(x_0+y_0)$.
 A: I recommend another approach. Let's use polar form to find $z_0$.
We see that $|z|=\sqrt{18^2+26^2}=\sqrt{1000}$. Thus $|z_0|$ is the cube root of that, namely $|z_0|=\sqrt{10}$.
To find the argument of $z_0$ (let's call it $\theta$), note that the argument of $z_0^3=z=18+26i$ is $3\theta$ and is given by $\tan3\theta=\frac{26}{18}=\frac{13}9$. Expanding,
$$\begin{align}
\frac{13}9 &= \tan 3\theta \\[2ex]
 &= \tan(2\theta+\theta) \\[2ex]
 &= \frac{\tan 2\theta+\tan\theta}{1-(\tan 2\theta)(\tan\theta)} \\[2ex]
 &= \frac{\frac{2\tan\theta}{1-\tan^2\theta}+\tan\theta}
   {1-\left(\frac{2\tan\theta}{1-\tan^2\theta}\right)(\tan\theta)} \\[2ex]
 &=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}
\end{align}$$
Letting $t=\tan\theta$ and rearranging that equation, we get
$$9t^3-39t^2-27t+13=0$$
Using the Rational Root Theorem we can factor that into
$$(3t-1)(3t^2-12t-13)=0$$
Which gives us the solutions
$$\tan\theta=t=\frac 13, \ \frac{6\pm5\sqrt 3}{3}$$
The smallest positive argument is given by the first solution, $\tan\theta=\frac 13$. We can quickly combine that with $|z_0|=\sqrt{10}$ to get the answer
$$z_0=3+i$$
This could have been seen by inspection much earlier, but I wanted to show a fuller exposition of getting that answer. Checking quickly shows us this is the desired cube root of $z$. In any case, we get
$$x_0=3, \ y_0=1$$
and the final answer

$$x_0y_0(x_0+y_0)=3\cdot 1(3+1)=12$$

