Simplifying an expression $\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$ if we know $x+y+z=0$ The following expression is given:
$$\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$$
Simplify it, knowing that $x+y+z=0$.

First $(x^3 + y^3 +z^3)(x+y+z) = x^4 + y^4 + z^4 + xy^3 + yx^3 + xz^3 +zx^3 +yz^3+y^3z = 0 $ which means 
$$ x^4 + y^4 + z^4 = -xy(x^2+y^2) - yz(y^2+z^2) -zx (z^2 + x^2) \ \ \text{(1)}$$
 $x+y+z=0$  also implies $x^2+y^2=z^2-2xy$, substitute all the sum of square we have
 $ x^4 + y^4 + z^4 = -xy(z^2-2xy) - yz(x^2-2xz)-zx(y^2-2zx)=-xyz(x+y+z)+2(x^2y^2+y^2z^2+z^2x^2) $ 
So 
$$ x^4 + y^4 + z^4=2(x^2y^2+y^2z^2+z^2x^2)  \ \ \text{(2)}$$
Next consider $ x^3+y^3-(x+y)^3 = -3xy(x+y) $ (this is a basic identity)
So we have $ x^3 + y^3 +z^3 = 3xyz $ for $x+y+z = 0$
 $3xyz(x^4+y^4+z^4)=(x^3 + y^3 +z^3)((x^4 + y^4 +z^4)=x^7+y^7+z^7+ x^4y^3 + y^4x^3 + x^4z^3 +z^4x^3 +y^4z^3+y^3z^4 = x^7+y^7+z^7 +x^3y^3(x+y)+y^3z^3(y+z)+z^3x^3(z+x)$  
By substitution with $\text{(1)}$ and $\text{(2)}$,
$$3xyz(x^4+y^4+z^4) =x^7+y^7+z^7-xyz(x^4+y^4+z^4)/2 $$
Therefore the fraction is $\frac{7}{2}$
Wow it's not this long in my thought. 
