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?

You can try eliminating the $a^3$ term, by multiplying the first equation by $b$ and the second equation by $\frac{a}{3}$ and subtracting the two.
$$ \begin{cases} 
  b \left(a^3 + 39 ab^2 - 18 = 0 \right) \\ 
  \frac{a}{3} \left( 3a^2 b + 13 b^3 - 5 = 0 \right)
\end{cases} \Rightarrow $$
$$ \begin{cases} 
  b a^3 + 39 ab^3 - 18b = 0  \\ 
   b a^3 + \frac{13}{3} a b^3 - \frac{5}{3} a = 0 
\end{cases} \Rightarrow \frac{104}{3} a b^3 + \frac{5}{3} a - 18 b = 0 $$
$$ a= \frac{54 b}{104 b^3 + 5} $$
Substituting $a$ into the second equation yields
$$ \frac{8748 b^3}{(104 b^3+5)^2} + 13 b^3 -5 =0$$ which is solved with $$b^3 = \frac{1}{8} \Rightarrow b = \frac{1}{2} $$
finally you get
$$ \begin{cases} a = \frac{3}{2} \\ b  = \frac{1}{2} \end{cases} $$
