<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] I found an answer from the comment.
\begin{align}
& \text{WLOG } a<b<c<d<e. \\
\ \\
& f(a)>0, f(b)<0, f(c)>0, f(d)<0, f(e)>0. \\
\ \\
& \therefore \text{ There exists 4 roots, one each  in } [a, b], [b, c], [c, d], [d, e] \qquad \text{by the Intermediate Value Theorem}.
\end{align}
</s>