<s>[INST] 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?
 [/INST] 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...
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