<s>[INST] Solving $a^3 - 33 ab^2 = -217$ and $3a^2 b - 11b^3 = 18$? I am looking for ways of solving systems like that:
$$\left\{\begin{array}{lcl} a^3 - 33 ab^2 = -217 \\ 3a^2 b - 11b^3 = 18 \end{array} \right.$$
I've tried turning it into a system of 2 equations with 4 variables but then, having the relations and 4 degrees of freedom makes it equally difficult. Are there any catches one could apply here?
 [/INST] Taking the resultant of $p=a^3-33ab^2+217$ and $q=3a^2b-11b^3-18$ with respect to $b$ immediately gives
$$
(4a^4 - 14a^3 + 111a^2 + 217a + 961)(4a^2 + 14a + 49)(2a^2 + 7a - 31)(2a - 7)=0.
$$
Hence the only rational solution is
$$
(a,b)=\left( \frac{7}{2},\frac{3}{2}\right).
$$
The quadratic equation $2a^2+7a-31=0$ yields
$$
(a,b)=\left( \frac{-7\pm 3\sqrt{33}}{2},\frac{-33\pm 7\sqrt{33}}{2}\right)
$$
Similarly we can solve the quadratic equation $4a^2 + 14a + 49=0$, which has non-real solutions. The quartic equation has four non-real roots.
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