Show product of $(r^5-r+2)$ over the 5 roots of $x^5+x^3+1=0$ is $1$ Let $(a,b,c,d,e)$ be the five roots of $x^5+x^3+1=0$, and let $g(x) = x^5-x+2$.
Prove 
$$
g(a)g(b)g(c)g(d)g(e) = 1
$$
preferably without pages of messy algebra.

We'll do some computations. As noted in G. Sassatelli's comment, for roots of our polynomial,
$$x^5-x+2 - (x^5+x^3+1)=-x^3-x+1.$$
Use polynomial division to get
$$x^2+1 + x^2(x^3+x-1)=x^5+x^3+1=0$$
and
$$-(x^3+x-1)=(1+x^2)/x^2,$$
since none of the roots are zero.
Realize that $abcde=-1$, so our product is equivalent to 
$$h(a)\cdots h(e)$$
with $h(x)=(1+x^2)$.
Use polynomial division again:
$$(1+x^2)x^3+1=x^5+x^3+1=0$$
$$(1+x^2)=-1/x^3$$
So our product is equivalent to
$$(-1)^5/(a^3b^3c^3d^3e^3)=1.$$
Please tell me if there is a sign error.

I believe there is a sort of euclidean algorithm behind all this here that will allow you to compute $g(a)\cdots g(e)$ for any $g$.
