jmc

Consider the polynomial $$\begin{array}{rl}p(x)& =\frac{a^5(x-b)(x-c)(x-d)(x-e)}{(a-b)(a-c)(a-d)(a-e)}+\frac{b^5(x-a)(x-c)(x-d)(x-e)}{(b-a)(b-c)(b-d)(b-e)}\\ & +\frac{c^5(x-a)(x-b)(x-d)(x-e)}{(c-a)(c-b)(c-d)(c-e)}+\frac{d^5(x-a)(x-b)(x-c)(x-e)}{(d-a)(d-b)(d-c)(d-e)}\\ & +\frac{e^5(x-a)(x-b)(x-c)(x-d)}{(e-a)(e-b)(e-c)(e-d)}.\end{array}$$Note that $p(x)$ is a polynomial of degree at most 4. Also, $p(a)=a^5,$ $p(b)=b^5,$ $p(c)=c^5,$ $p(d)=d^5,$ and $p(e)=e^5.$ This might lead us to conclude that $p(x)=x^5,$ but as we just observed, $p(x)$ is a polynomial of degree 4.

So, consider the polynomial $$q(x)=x^5-p(x).$$The polynomial $q(x)$ becomes 0 at $x=a,$ $b,$ $c,$ $d,$ and $e.$ Therefore, $$q(x)=x^5-p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)r(x)$$for some polynomial $r(x).$

Since $p(x)$ is a polynomial of degree at most 4, $q(x)=x^5-p(x)$ is a polynomial of degree 5. Furthermore, the leading coefficient is 1. Therefore, $r(x)=1,$ and $$q(x)=x^5-p(x)=(x-a)(x-b)(x-c)(x-d)(x-e).$$Then $$p(x)=x^5-(x-a)(x-b)(x-c)(x-d)(x-e),$$which expands as $$p(x)=(a+b+c+d+e)x^4+\cdots .$$This is important, because the expression given in the problem is the coefficient of $x^4$ in $p(x).$ Hence, the expression given in the problem is equal to $a+b+c+d+e.$ By Vieta's formulas, this is $\boxed{-7}.$