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

$$ \begin{cases} a = \frac{3}{2} \\ b  = \frac{1}{2} \end{cases} $$
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 $3a^2$.