<s>[INST] How to solve $a^3 + 39 ab^2 - 18 = 0$, $3a^2 b + 13 b^3 - 5 = 0$ In this answer, the user @123 has claimed that by solving the system
$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$
we give $ a = \dfrac 32$ and $ b = \dfrac12$. Could anyone explain for me that how one can solve such a system, please?
 [/INST] Here is one approach. Note from the second equation you have
$$
3a^2 = 5/b - 13b^2
$$
which we can substitute into the first to get
$$
18 = a\left(a^2+39b^2\right)
   = a\left(\frac{5}{3b} - \frac{13b^2}{3}+39b^2\right)
$$
so
$$
3a^2 = \left(\frac{18\cdot 3}{\frac{5}{3b} - \frac{13b^2}{3}+39b^2} \right)^2
$$
which implies an equation in $b$,
$$
\frac{5}{b} - 13b^2 = \left(\frac{18\cdot 3}{\frac{5}{3b} - \frac{13b^2}{3}+39b^2} \right)^2,
$$
since both sides are equal to $3a^2$.
</s>