<s>[INST] Simplifying $\frac{x^4(y-z)+y^4(z-x)+z^4(x-y)}{(y+z)^2+(z+x)^2+(x+y)^2}$ I need help to simplify this expression.
$$\frac{x^4(y-z)+y^4(z-x)+z^4(x-y)}{(y+z)^2+(z+x)^2+(x+y)^2}$$
I only know to start, I extracted the $(x-y)$ from top one, and then I don't know what's next. Extracting $(x-y)$ from the numerator:
$$(x^4y - y^4x) - (zy^4-zx^4)+z^4(x-y)$$
$$ = xy(x^3-y^3)- z(x^4-y^4)+z^4(x-y)$$
$$ = xy(x-y)(x^2+xy+y^2)-z(x-y)(x+y)(x^2+y^2)+z^4(x-y)$$
$$ = (x-y)\left[ xy(x^2+xy+y^2) - z(x+y)(x^2+y^2) +z^4 \right]$$
$$ = (x-y)(x^3y+x^2y^2+xy^3-zx^3-zy^3+z^4-zxy^2-zx^2y)$$
 [/INST] The nominator is a degree 5 polynomial and the denominator degree 2 polynomial, so if the fraction should simplify into a polynomial, it would have to be of degree 3, symmetric, and with the same number of positive and negative terms. My first guess is to consider the symmetric polynomial
$$
  (x-y)(y-z)(z-x) 
= y (z^2-x^2) + z (x^2-y^2) + x (y^2-z^2). 
$$
Let's try when we multiply it by the denominator of the fraction:
\begin{align}
  &\frac 1 2((x-y)(y-z)(z-x) 
  \Big((y+z)^2 + (z+x)^2 + (x+y)^2\Big) \\
&= \Big(y (z^2-x^2) + z (x^2-y^2) + x (y^2-z^2)\Big) \Big(x^2 + y^2 + z^2 + xy + yz + zx\Big)\\ 
&= \Big(y (z^2-x^2) + \ldots \Big) \Big(x^2 + y^2 + z^2 + xy + yz + zx\Big)\\ 
&= \Big(y (z^2-x^2) + \ldots\Big) \Big(x^2 + y^2 + z^2 \Big) \quad \color{red}{=: A}\\
& \ + \Big(y (z^2-x^2) + \ldots\Big) \Big(xy + yz + zx\Big)  \quad \color{red}{=: B}, 
\end{align}
where I use the notation $(\it \text{expression}+\ldots)$ to represent the cyclical completion of the expression $x\to y \to z (\to x)$. For example $(xy+\ldots)$ represents the expression $xy + yz + zx$, which could equivalently be represented by $(yz+\ldots)$ or $(yz+\ldots)$.

Taking into account that the expression $\big(x^2 + y^2 + z^2\big)$ is symmetric, we can write
\begin{align}
A &= \Big( y (z^2-x^2) \big(x^2 + y^2 + z^2 \big) + \ldots \Big) \\
&=\Big( y \big(z^2x^2- x^4 + z^2 y^2 -x^2 y^2 + z^4 - x^2 z^2\big) + \ldots  \Big) \\
&= \Big(2y (z^2 y^2 -x^2 y^2 + z^4-x^4) + \ldots  \Big) \\
&= \Big(y^3 z^2 - x^2 y^3 + \ldots  \Big) + 2\Big(y(z^4-x^4) + \ldots  \Big) \\
&= \Big(x^3 y^2 - x^2 y^3 +\ldots \Big)
 -\Big(x^4(y-z) + y^4(z-x) + z^4 (x-y) \Big),
\end{align}
where in the last line we took into the fact that the cyclical term $(y^3z^2+\ldots) = (x^3y^2+\ldots)$.

Taking into account that the expression $\big(xy + yz + zx\big)$ is symmetric, we can write
\begin{align}
B &= \Big(y (z^2-x^2) \big(xy + yz + zx\big) + \ldots \Big) \\
&\Big(
xy^2z^2 - x^3 y^2 + y^2 z^3 - x^2 y^2 z + xyz^3 - x^3 y z + \ldots \Big) \\
&=xyz \Big(yz - xy + z^2 - x^2 + \ldots \Big)+
 \Big(- x^3 y^2 + y^2 z^3 +  \ldots \Big) \\
&= \Big(- x^3 y^2 + y^2 z^3 +  \ldots \Big) \\
&= \Big(- x^3 y^2 + x^2 y^3 +  \ldots \Big) 
\end{align}

We conclude that
$A + B = -\Big(x^4(y-z) + y^4(z-x) + z^4 (x-y) \Big),$
and so
$$
\frac{x^4(y-z) + y^4(z-x) + z^4 (x-y)}
{(y+z)^2 + (z+x)^2 + (x+y)^2} =
  - \frac 1 2 (x-y)(y-z)(z-x).
$$
</s>