Find $x$ and $y$ such that $(x+3i)(x+iy)=1-i$ I tried expanding the brackets, equating Real and Imaginary parts and finally substituting $x$ in terms of $y$. I ended up with the equation:
$$3y^3 + 19y^2 + 33y +8=0$$
Using a calculator, I got a root at $y\simeq-0.288$ which is approx. $\frac{-7}{25}$.
I should be getting $y=\frac{-7}{25}$ and $x=\frac{1}{25}$. How should I go about finding $x$ and $y$ without a calculator?
 A: Expanding the left hand side and comparing real and imaginary parts yields
\begin{eqnarray*}
x^2-3y&=&\hphantom{-}1\tag{1}\\
xy+3x&=&-1\tag{2}
\end{eqnarray*}
The first equation shows that $y=\tfrac{x^2-1}{3}$ and substituting into the second yields
$$-1=x\frac{x^2-1}{3}+3x=\frac13x^3+\frac{8}{3}x,$$
and so $x$ is a root of the cubic equation
$$X^3+8X+3=0.\tag{3}$$
By the rational root test, this cubic has no rational roots. From the second equation we find that $x\neq0$ and $y=-\frac1x-3$, which shows that also $y$ cannot be rational. In particular, the solutions you give cannot actually be solutions.
The roots of $(3)$ can be expressed in terms of radicals by means of the cubic formula. You may find that the cubic has precisely one real root because its discriminant is negative. It does not simplify to anything 'nice'.
The only real solution of the equation $(3)$ is
$x=\sqrt[3]{-\dfrac32+\dfrac1{18}\sqrt{6873}}-\sqrt[3]{\dfrac32+\dfrac1{18}\sqrt{6873}}$
Moreover ,
$y=\dfrac13\sqrt[3]{\dfrac{1267}{54}+\dfrac16\sqrt{6873}}+ \dfrac13\sqrt[3]{\dfrac{1267}{54}-\dfrac16\sqrt{6873}}-\dfrac{19}9$
