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?

Setting $a=bt$ then we get
$$b^3+t^3+39b^3t-18=0$$
$$3b^3t^2+13b^3-5=0$$ eliminating $b^3$ then we get
$$5t^3-54t^2+195t-234=0$$
one solution is $t=3$ then you will get $a=3b$
plugging this in the given equation we get $$a=\frac{3}{2},b=\frac{1}{2}$$
