<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] I like Pavel's answer, but here's a slightly different way.
Starting with $(x-y)(x^3y+x^2y^2+xy^3−zx^3−zy^3+z^4−zxy^2−zx^2y)$, rearrange the longer expression to:
$$(x^3y-zx^3) + (xy^3-zxy^2) + (x^2y^2 - zx^2y) + (z^4 - zy^3)$$
$$ = (y-z)(x^3 + xy^2 + x^2y) - z(z^3-y^3)$$
$$ = (y-z)(x^2y + xy^2 + x^2y -z(y^2+yz+z^2)) $$
Giving us $(x-y)(y-z)(x^2y+xy^2+x^2y-y^2z-yz^2-z^3)$ as the numerator. Again, rearrange the new long expression as:
$$(xy^2-zy^2) + (x^3-z^3) + (x^2y-yz^2)$$
$$ = (x-z)y^2 + (x-z)(x^2+xz+z^2)+y(x-z)(x+z)$$
And we can factor out the $(x-z)$. This gives the numerator as:
$$(x-y)(y-z)(x-z)(x^2+y^2+z^2+xy+xz+yz)$$
But if we expand the denominator, we get:
$$x^2+2xy+y^2 + x^2+2xz+z^2 + y^2+2yz+z^2 = 2(x^2+y^2+x^2+xy+xz+yz)$$
And we recognize that as having the same expression in parentheses as the last expression in the numerator! So when we divide we get:
$$\frac{(x-y)(y-z)(x-z)}{2} = - \frac{(x-y)(y-z)(z-x)}{2}$$
Which is the same as Pavel's answer. Taking the negative out just makes the numerator look symmetrical. If you expand the numerator, you get another symmetrical expression:
$$\frac12 \big(xy(x-y) + yz(y-z) + zx(z-x) \big)$$
which is longer but, I think, just a bit nicer-looking.
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