We know $x$ and $y$ are roots of $$ax^2+ bx + c= 0$$ and $$P = xy , S = x + y $$ How we can calculate $A$ versus $S$ and $P$ ?
$A = x^3 + {1\over x^3}$
My try : I found a complex and long relation by this equation $$x^2-Sx+p=0$$ but I want a simply relation.
On
There is no way to write $A$ as a function of $S$ and $P$ (and no other variables). Indeed, notice that the definitions of $S$ and $P$ do not change if you swap which root you're calling $x$ and which root you're calling $y$. If you could write $A$ in terms of $S$ and $P$, the same would be true of $A$. But this is not true: if you swap $x$ and $y$ then $A$ turns into $y^3+\frac{1}{y^3}$, which is not the same as $A=x^3+\frac{1}{x^3}$ (at least for most values of $x$ and $y$).
On
The problem is equivalent to eliminating $x$ between the equations:
$$ \begin{cases} x^2 - S x + P = 0 \\[1ex] x^6 - A x^3 + 1 = 0 \end{cases} $$
which amounts to the resultant of the two polynomials in $x$ being $0$:
$$P^3 \cdot A^2 + (3 P^4 S - P^3 S^3 + 3 P S - S^3) \cdot A + (P^6 - 2 P^3 + 9 P^2 S^2 - 6 P S^4 + S^6 + 1) = 0$$
The above is a quadratic in $A$ with coefficients depending on $P, S$ which can be solved for $A$, giving (in general) two values corresponding to the two roots of the original quadratic.
Let $e$ and $f$ be the roots of $ax^2+bx+c=0$. This implies that: $$P = ef \, \text{ and } \, S = e+ f$$ Also, $$e-f = \sqrt{(e+f)^2-4ef} = \sqrt{ S^2-4P}$$ $$\implies e = \frac{1}{2}[S + \sqrt{S^2-4P}]\, \text{ and } \, f = \frac12[S - \sqrt{S^2-4P}]$$
We want $A = x^3 + \frac{1}{x^3}$. To do this, note that we can write: $$ax^2+bx + c = (x-e)(x-f)$$ Hence, $$A = e^3 + \frac{1}{e^3} \, \text{ or } \, f^3 + \frac{1}{f^3}$$ because $x = e \, \text{ or } \, f$ satisfies the quadratic.
Hence, $$A = \frac{(S \pm \sqrt{S^2-4P})^3}{8} + \frac{8}{(S \pm \sqrt{S^2-4P})^3}$$ incorporating the $(+)$ sign for $e$ and the $(-)$ sign for $f$.
Hence, we have: $$\boxed{ A = \frac{(S \pm \sqrt{S^2-4P})^6 + 2^6}{(2S \pm 2\sqrt{S^2-4P})^3}}$$