Solving the system $a^3 + 15ab^2 = 9$, $\;\frac 35 a^2b + b^3 = \frac 45$ I have problem solving the following system of two cubic equations.
$$\begin{cases}
a^3 + 15ab^2 = 9 \\
\frac 35 a^2b + b^3 = \frac 45
\end{cases}
$$
I don't have any idea how to approach this kind of problem.
I'm looking for solutions included in the set of real numbers (no complex solutions).
Is there any general solution to the problem?

Starting from
$$\begin{align}
a^3 + 15ab^2 = 9 \\
\frac 35 a^2b + b^3 = \frac 45
\end{align}$$
let $b=ax$ as obviously $b\neq0$:
$$\begin{align}
a^3 + 15x^2a^3 = 9 \\
3 xa^3 + 5x^3a^3 = 4
\end{align}$$
dividing both sides eliminates $a$:
$$
\frac{1+ 15x^2}{3 x + 5x^3} = \frac 94
$$
so you have a cubic in $x$, for which there are explicit solutions.
