<s>[INST] Solve the equation $z^{3}=18+26i$, where $z=x+yi$ and $x,y$ are integers. $(x+yi)^{3}=(x^{3}\text{−}3xy^{2})+(3x^{2}y\text{−}y^{3})i=18+26i$
  which leads to $$\begin{cases}
x^{3}\text{−}3xy^{2} & =18\\
3x^{2}y\text{−}y^{3} & =26
\end{cases}$$
The problem here is that I don't know to precede further. 
I've been told to set $y=tx$ in the equatlity $18(3x^{2}y\text{−}y^{3})=26(x^{3}\text{−}3xy^{2})$ and then if $x\neq0$ and $y\neq0$ then $18(3t\text{−}t^{3})=26(1\text{−}3t^{2})$ which is equivalent to $$(3t−1)(3t^{2}-12t-13)=0$$. The only rational solution of this equation is $t=\frac{1}{3}$; hence, $x=3,y=1$, and $z=3+i$.
But I don't understand that solution (the substitution $y=tx$ to be specific) and how they ended up with $18(3t\text{−}t^{3})=26(1\text{−}3t^{2})$ from that substitution?
 [/INST] $$x^{3}\text{−}3xy^{2} =18 \Rightarrow x(x^2-3y^2) = 18.$$
This means that $x$ is a divisor of $18$, i.e. $x \in \{1,2,3,6,9,18\}$.
Similarly:
$$3x^{2}y\text{−}y^{3} =26 \Rightarrow y(3x^2-y^2) = 26,$$
which means that $y$ is a divisor of $26$, i.e. $y \in \{1,2,13,26\}$.
Moreover, we know that:
$$|z^3|^2 = 18^2 + 26^2 = 1000.$$
Then, $|z|^2 = 1000^{\frac{1}{3}} = 10 = x^2 + y^2$
We can reduce the sets for $x$ and $y$, since $x^2 < 10$ and $y^2 < 10$. We get:
$$\begin{cases}x \in \{1,2,3\}\\
y \in \{1,2\}
\end{cases}$$
It is clear now that the only solution is $x= 3$ and $y=1$.
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