algebra precalculus - Compute $ax^3+by^3$ given $ax^2+by^2$ and $ax+by$.

Given are positive real numbers $A$ and $B$ and positive integers $a$ and $b$ such that
$$ \begin{aligned} ax+by &= A\\ ax^2+by^2 &= B.\end{aligned} \tag{*}$$
What are the possible values of

$$ ax^3 + by^3 = C?$$

Think of it as $a+b$ numbers, $a$ of them are $x$ and $b$ of them are $y$, such that the sum and the sum of their squares are given and the sum of their cubes is asked. I'm not sure if the answer can be expressed as a rational function of $A$, $B$, $a$ and $b$, but I hope so.

Note that one has the equation
$$ (A-ax)^2 = (by)^2 = b(B-ax^2),$$
thus
$$\begin{aligned} a(a+b)x^2 &= bB-A^2 + 2ax,\\ b(a+b)y^2 &= aB-A^2 + 2by.\end{aligned} \tag{1}$$
Multiplying with $x$ and adding gives
$$\begin{aligned} (a+b)C &= B(bx+ay)-A^2(x+y) + 2B \\ &= B(a+b)(x+y)- B(ax+by) - A^2(x+y)+2B \\&= \big(B(a+b)- A^2\big)\,(x+y) - AB+ 2B. \end{aligned}$$

So this boils down to determining possible values of $x+y$. Here I am a bit stuck. Note also that from (*) one can deduce (multiply the first equation with $x$, $y$, add and subtract the second):
$$ (a+b)xy = A(x+y) - B.$$

One could simply solve the quadratic equations (1), verify what combination of solutions works for the original problem and use this to calculate $C$. But actually I am interested in a solution that will work for higher order cases as well, (3 unknowns, sum of first, second and thirth powers is known and the sum of fourth powers must be determined) so I am looking for an approach avoids this. Maybe there is a quadratic equation with $x$ and $y$ as roots hidden somewhere?

Answer

Well, you have 3 equations,

$$ax+by = A\tag{1}$$

$$ax^2+by^2 = B\tag{2}$$

$$ax^3+by^3 = C\tag{3}$$

If you want to express C in terms of $a,b,A,B$, you can eliminate $x,y$ between $(1), (2), (3)$ to have an equation purely in $a,b,A,B,C$ (easily done in Mathematica using the Resultant command).

However, we get a quadratic in C. But we can use (1) and (2) to simplify its discriminant D and I ended up with,

$$C = \frac{A(-2A^2+3B(a+b))\pm(a-b)(a b)D^3}{(a+b)^2}$$

where,

$$D= (x-y) = \sqrt{\frac{-A^2+(a+b)B}{ab}}\tag{4}$$

Hence, C is not a rational function of $a,b,A,B$, as you need to take square roots as shown by $(4)$.