Finding the value of an expression using the roots of a given polynomial 
Let $a,$ $b,$ $c,$ $d,$ and $e$ be the distinct roots of the equation $x^5 + 7x^4 - 2 = 0.$ Find
\begin{align*} &\frac{a^4}{(a - b)(a - c)(a - d)(a - e)} + \frac{b^4}{(b - a)(b - c)(b - d)(b - e)} \\ &\quad + \frac{c^4}{(c - a)(c - b)(c - d)(c - e)} + \frac{d^4}{(d - a)(d - b)(d - c)(d - e)} \\ &\quad + \frac{e^4}{(e - a)(e - b)(e - c)(e - d)}. \end{align*}

I started by using Vieta's formulas to obtain the following system of equations:
\begin{cases}
&\sum_{\text{cyc}} a = -7, \\
&\sum_{\text{cyc}} ab = 0, \\
&\sum_{\text{cyc}} abc = 0, \\
&\sum_{\text{cyc}} abcd = 0, \\
&abcde = 2.
\end{cases}
Now let $f_a(x) = (x - b)(x - c)(x - d)(x - e)$, then $f_a(x) = x^4 - x^3\sum_{\text{cyc}} b + x^2 \sum_{\text{cyc}} bc -x \sum_{\text{cyc}}bcd + bcde$. By using the above system of equations, we can eventually simplify it down and see that $f_a(a) = a^4 + (7a^3 + a^4)4$. Now there are a ton of options for what we can do, but now the sum we desire to find is given by $$\sum_{\text{cyc}} \frac{a^4}{a^4 + (7a^3 + a^4)4}.$$
Using that $a^5 + 7a^4 - 2 = 0$, we can simplify further to find that $f_a(a)a = 10 - \frac{14}{a+7}$ and $\frac{2a}{a+7} = a^5$ so $$\sum_{\text{cyc}} \frac{a^4}{a^4 + (7a^3 + a^4)4} \cong \sum_{\text{cyc}} \frac{a}{5a+28}.$$
Now this should be simpler, but after thinking about it I don't really see a good way to proceed. What can I do? Did I drive into the weeds, or am I still on the misty road and I just don't see the end?
 A: a Mobius transformation takes polynomials into rational functions. The roots of the polynomial are mapped to the roots of the resulting numerator.   with polynomial $f(x) = x^5 + 7 x^4 - 2,$  I then wrote
$$x =  \frac{28t}{-5t+1}  $$
The resulting rational function $f(x)$ in $t$   was
$$  g(t) = \frac{-4296342t^5 + 4296342t^4 + 2500t^3 - 500t^2 + 50t - 2}{-3125t^5 + 3125t^4 - 1250t^3 + 250t^2 - 25t + 1}   $$
Now, $g(t)$ is zero when
$$  h(t) = -4296342t^5 + 4296342t^4 + 2500t^3 - 500t^2 + 50t - 2  $$
is zero.
If   $t$  is a root of $h(t) $  and
$$t =  \frac{x}{5x+28} , $$
then $x$   is a root of your $x^5 + 7 x^4 - 2.$  The sum of the five roots of $h$  gives the sum of your desired $  \frac{x}{5x+28}  $  with $x$  the five roots...
To get monic, let $h(t) = -4296342 h_1(t),$   so that
$$  h_1(t) \; = \; \; t^5 \; - \; t^4 \;  - \;\frac{1250t^3}{2148171}  + \; \frac{250t^2}{2148171}   - \; \frac{25t}{2148171}   + \; \frac{1}{2148171} $$
and the sum of the roots of $h_1$ is ...
A: Hint: Turns out Vieta isn't needed. Irrespective of which polynomial they are roots of, as long as $a,b,c,d,e$ are distinct,
$$\sum_{cyc} \frac{a^4}{(a-b)(a-c)(a-d)(a-e)}=1$$
Will expand on the hint in case you can't complete the proof...
