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$(x+yi)^{3}=(x^{3}\text{−}3xy^{2})+(3x^{2}y\text{−}y^{3})i=18+26i$ 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?

mohamez
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3 Answers3

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$(x^2 + y^2)^3 = |z^3| = 18^2 + 26^2 = 1000$, so $x^2 + y^2 = 10$. There are only 8 possibilities for that: $(\pm 1, \pm 3)$ and $(\pm 3, \pm 1)$. One works.

Robert Israel
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You can solve the original equation in this way:

$$z^3=18+26i=27+3\cdot 9i+3\cdot3i^2+i^3=(3+i)^3\\z=3+i$$Which is a solution, and to check that the other 2 solutions aren't, just multiply by the 2 other unit roots of $3+i$.

76david76
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$$x^{3}\text{−}3xy^{2} =18 \Rightarrow x(x^2-3y^2) = 18.$$ 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 $y=1$.

the_candyman
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