$\begin{array}{l}\text { 6. Given } P(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+ \\ a_{3} x^{2}+a_{4} x+a_{5} \text {, and when } k=1,2,3,4 \text {, } P(k) \\ =k \cdot 1997 \text {. Then } P(10)-P(-5)=\end{array}$
6.75315 .

Let $Q(x)=P(x)-1997 x$. Then for $k=1,2,3,4$, $Q(k)=P(k)-1997 k=0$. Hence, $1,2,3,4$ are roots of $Q(x)=0$. Since $Q(x)$ is a fifth-degree polynomial, we can assume
$$
Q(x)=(x-1)(x-2)(x-3)(x-4)(x-r) \text {. }
$$
Therefore,
$$
\begin{array}{l} 
P(10)= Q(10)+1997 \cdot 10 \\
= 9 \cdot 8 \cdot 7 \cdot 6(10-r)+1997 \cdot 10 . \\
P(-5)= Q(-5)+1997(-5) \\
=(-6)(-7)(-8)(-9)(-5-r) \\
+1997 \cdot(-5) . \\
\therefore P(10)-P(-5) \\
= 9 \cdot 8 \cdot 7 \cdot 6 \cdot 15+1997 \cdot 15=75315 .
\end{array}
$$