Girard Symmetry

$x^2+xy+y^2=0 \\=> (\dfrac{x}{y})^2+\dfrac{x}{y}+1=0$

Let , $\frac{x}{y}=a$ then $a^2+a+1=0$ whose roots are $\omega,\omega^2$ (these are cube roots of unity)

Now come to the required expression $\dfrac{x^{2001}+y^{2001}}{(x+y)^{2001}} \\ =\dfrac{(x/y)^{2001}+1}{(x/y+1)^{2001}} \\ =\dfrac{a^{2001}+1}{(a+1)^{2001}}$

Now put either $\omega$ or $\omega^2$ in place of $a$ it becomes $\dfrac{\omega^{2001}+1}{(\omega+1)^{2001}}$

Using the formula $\omega^2+\omega+1=0$ we get the expression will be $\boxed{-2}$

(Throughout this solution, whenever $P_n$ is used, it implies the nth power sum)

A curious fact is that $\frac{x}{x+y}$ and $\frac{y}{x+y}$ are the two solutions of the polynomial $t^2-t+1=0$ .

We need to find $P_{2001}$ for this polynomial.

We see that $a_2=1, a_1=-1$ and $a_0=1$ .

So we form the recursion, $1.P_{k+2}+(-1).P_{k+1}+1.P_k=0$ (This recursion is valid as a result of Newton-Girard formula)

or, $P_ {k+2}=P_{k+1}-P_k$ or, $P_ {k+2}=P_k-P_{k-1}-P_k$ So, $P_ {k+2}=-P_{k-1}$ Implies that, $P_{2001}=-P_{1998}=P_{1995}=...=P_{3}$

We are supposed to find $\dfrac {x^{3}+y^{3}}{(x+y)^{3}}$ .

That much math is left to the reader. The answer you'll land on is bound to be $\boxed{-2}$ if you do not make any wrong moves.