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 $x+y+z=0$.
 A: Note: Using Newton's identities, we can calculate the below expressions more easily, following the easy recursive definition.
But, your idea of writing as roots of third degree polynomial works I believe, but requires some work and we show that here:
Let $\displaystyle x,y,z$ be roots of $\displaystyle t^3 + at - b = 0$. We have that $\displaystyle a = xy+yz+zx$ and $\displaystyle b = xyz$.
Since $\displaystyle t^3 = b - at$, multiply by $\displaystyle t^4$ we get $\displaystyle t^7= bt^4 - a t^5$.
Setting $\displaystyle t=x,y,z$ in turn and adding gives us

$\displaystyle x^7 + y^7 + z^7 = b(x^4 + y^4 + z^4) - a(x^5 + y^5 + z^5)$

Similar to above, we get $\displaystyle t^5 = bt^2 - a t^3$, setting $\displaystyle t=x,y,z$ and adding gives us

$\displaystyle x^5 + y^5 + z^5 = b(x^2 + y^2 + z^2) - a(x^2 + y^3 + z^3)$.

Similarly we get

$\displaystyle x^3 + y^3 + z^3 = 3b$

We also have $\displaystyle (x+y+z)^2 = 0$, giving us

$\displaystyle x^2 + y^2 + z^2 = -2a$.

Thus

$\displaystyle x^5 + y^5 + z^5 = -2ab - 3ab = -5ab$.

Now $\displaystyle t^4 = bt - at^2$ and in a similar fashion we get

$\displaystyle x^4 + y^4 + z^4 = -a(x^2+y^2 +z^2) = 2a^2$
Thus $\displaystyle xyz(x^4 + y^4 + z^4) = 2a^2 b$
and $\displaystyle x^7 + y^7 + z^7 = 2a^2b - (-5a^2b) = 7a^2 b$.
Thus the given expression is $\displaystyle \frac{7}{2}$.

This approach can be used to generate identities.
For instance, show that

$\displaystyle 10(x^7 + y^7 + z^7) = 7(x^2 + y^2 + z^2)(x^5 + y^5 + z^5)$

A: Exploit the innate symmetry! Using Newton's identities  to rewrite the power sums as elementary symmetric functions is very simple because $\rm\:e_1 = x+y+z = 0\:$ kills many terms.
Write $\rm\ \ c = e_2 = xy + yz + zx,\ \ \ d = e_3 = xyz,\ \ \ p_k =\: x^k + y^k + z^k.$
$\rm\qquad\qquad p_1\ =\ e_1 = 0$
$\rm\qquad\qquad p_2\ = {-}2\: c$
$\rm\qquad\qquad p_3\ =\ \ \ 3\: d$
$\rm\qquad\qquad p_4\ =  - c\: p_2\ =\ 2\: c^2$
$\rm\qquad\qquad p_5\ = -c\: p_3 +\: d\: p_2\ = {-}5\: c\: d$
$\rm\qquad\qquad p_7\ = -c\: p_5 +\: d\: p_4\ =\ 7\: c^2 d$
Hence $\rm\displaystyle\ \frac{p_7}{p_4 d}\: =\: \frac{7\: c^2\: d}{2\: c^2\: d}\: =\: \frac{7}{2}$
A: Use that $z=-(x+y)$, so you have that the numerator turns out to be $$x^7+y^7-(x+y)^7=-7x^6y-21x^5y^2-35x^4y^3-35x^3y^4-21x^2y^5-7xy^6$$Also, the denominator turns out to be $$xy(-x-y)(x^4+y^4+(-x-y))^4=-2x^6y-6x^5y^2-10x^4y^3-10x^3y^4-6x^2t^5-2xy^6$$You can factor a $7$ from the numerator and a $2$ from the denominator, and the answer turns out to be $$\frac{7}{2}$$
A: 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. 
