Question 12
Let $a, b, c, d, e$ be distinct real numbers. Prove that the equation
$$
\begin{aligned}
& (x-a)(x-b)(x-c)(x-d) \\
& +(x-a)(x-b)(x-c)(x-e) \\
& +(x-a)(x-b)(x-d)(x-e) \\
& +(x-a)(x-c)(x-d)(x-e) \\
& +(x-b)(x-c)(x-d)(x-e)=0
\end{aligned}
$$
has 4 distinct real solutions.
$$
\begin{aligned}
& (x-a)(x-b)(x-c)(x-d) \\
& +(x-a)(x-b)(x-c)(x-e) \\
& +(x-a)(x-b)(x-d)(x-e) \\
& +(x-a)(x-c)(x-d)(x-e) \\
& +(x-b)(x-c)(x-d)(x-e)=0
\end{aligned}
$$
has 4 distinct real solutions.