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?

Disclaimer: This answer is probably not appropriate considering (algebra-precalculus) tag, but I'll write it anyway since it might be useful to others stumbling upon this question.
The system is easier to solve when considering original equation $$(a+b\sqrt{13})^3=18+5\sqrt{13}.$$
If we look at norm $N(x+y\sqrt{13}) = x^2-13y^2$, it turns out that $N(18+5\sqrt{13}) = -1$. By multiplicativity of $N$ it follows that $N(a+b\sqrt{13}) = -1$, which gives us $a^2-13b^2 = -1$. Substitute $13b^2 = a^2 + 1$ in $a^3+39ab^2 - 18 = 0$ to get equation $$4a^3+3a-18 = 0.$$
By rational root theorem, the only rational root of the last equation is $a=\frac 32$. It follows that $13b^2 = (3/2)^2+1$, or $b^2 = \frac 14$. From the equation $3a^2b + 13b^3 = 5$, we know that $b$ must be positive, so $b = \frac 12$.
Note that there are complex solutions as well.
