<s>[INST] $a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive 4 real roots. 
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive four real roots.


$$(x-a)(x-b)(x-c)(x-d) \\ +(x-a)(x-b)(x-c)(x-e) \\ +(x-a)(x-b)(x-d)(x-e) \\ +(x-a)(x-c)(x-d)(x-e) \\ + (x-b)(x-c)(x-d)(x-e) =0.$$

My attempt:
\begin{align}
& f(x)=\sum_{cyc} (x-a)(x-b)(x-c)(x-d). \\
& f(a)=(a-b)(a-c)(a-d)(a-e). \\
& f(b)=(b-a)(b-c)(b-d)(b-e). \\
& f(c)=(c-a)(c-b)(c-d)(c-e). \\
& f(d)=(d-a)(d-b)(d-c)(d-e). \\
& f(e)=(e-a)(e-b)(e-c)(e-d). \\
\ \\
& f(0)=\sum_{cyc} abcd.\\
\end{align}
I put some arbitrary integers in $a, b, c, d, e$ and drew a graph of $\displaystyle f(x)=\sum_{cyc} (x-a)(x-b)(x-c)(x-d).$
It sure makes 4 distinct roots... How can we prove this?

Values in the graph above:
$a=-1.34, b=-4.67, c=-2.91, d=0.33, e=-6.09$
 [/INST] Let $F(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$.  Then your polynomial is $f(x)=F'(x)$. Because $a,b,c,d,e$ are all distinct, by Rolle's theorem, between any two consecutive roots of $F(x)$ will be a root of $f(x)$.   Thus, $f(x)$ will have 4 distinct real roots because $F(x)$ has 5 distinct real roots.
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