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?
 A: Let's convert this to a nice single-variable cubic. Let $a = kb$. Then we have
$$
b^3(k^3 + 15k) = 9
\\ b^3(3k^2 + 5) = 4
$$
Now multiply top by $4$, bottom by $9$, and subtract to get
$$
b^3(4k^3 - 27k^2 + 60k - 45) = 0.
$$
We definitely don't have $b = 0$, so it's the cubic in $k$ that must be $0$. The discriminant is negative, so there's only the one real root, and it's clearly positive. By the rational roots theorem, if $k$ is rational, its numerator will divide $45$ and its denominator with divide $4$. We can then do some divisibility checks to rule out some of the possibilities. Since all coefficients except $k^3$ are divisible by $3$, $k$'s numerator must be divisible by $3$. But $k$'s numerator can't be divisible by $9$, because $4k^3 -27k^2 + 60k$'s numerator is divisible by $27$ but $45$ isn't. So the numerator is either $3$ or $15$. At this point we can just try all 6 remaining possibilities and find $k= 3$ works. So $k = 3$, and we have
$$
b^3(27+45) = 9\Longrightarrow b^3 = \frac{9}{72}\Longrightarrow b = \frac{1}{2}.
$$
So the solution is
$$
(a,b) = (3/2,1/2)
$$
A: 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.
