Solving $a^3 - 33 ab^2 = -217$ and $3a^2 b - 11b^3 = 18$?

Question

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?

Answer 1

Does this help? $$(a-\sqrt{11} i b)^3 = a^3-3\sqrt{11}i a^2 b -33ab^2+11\sqrt{11}ib^3$$ $$=(a^3 -33ab^2)-\sqrt{11}i(3 a^2 b-11b^3)$$ $$=-217-\sqrt{11} \, 18 \, i$$

Answer 2

This is one of the other approaches you would like to try.

Multiply the first equation by $18$, multiply the second one by $217$, then add them together, we get:

$18a^3-18*33ab^2+217*3a^2b-217*11b^3=0$

$\Leftrightarrow 18a^3+651a^2b-594ab^2-2387b^3=0$

The last equation is homogeneous of degree 3, so you could solve the associate cubic equation to compute $a$ by $b$ (Luckily this cubic equation has a rational root).

Answer 3

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.