calculus - Show that $a - b mid f(a) - f(b)$

I have seen this lemma elsewhere.

Suppose $A$ is a domain, and $f \in A[X]$. Prove that

$$a - b \mid f(a) - f(b)$$

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I need to prove this.

$$f(a) - f(b) \equiv 0 \pmod{a-b}$$ basically.

Let, $a - b = c$

$$f(a) - f(b)/(a-b) = f'(\xi)$$ for Some $\xi \in (a, b)$.

But I dont see it showing divisibility

Answer

Let $A$ be a domain with field of fractions $K$ and let $$ f(X)=a_0+a_1X+a_2X^2+\cdots+a_nX^n $$ be a polynomial in $A[X]$. Then for all $a\neq b\in A$, $$ \begin{eqnarray} f(b)-f(a)&=&(a_0+a_1b+\cdots+a_nb^n)-(a_0+a_1a+\cdots+a_na^n)\\ &=& (b-a)\left(a_1+a_2\frac{b^2-a^2}{b-a}+\cdots+a_n\frac{b^n-a^n}{b-a}\right) \end{eqnarray} $$ is certainly valid in $K$. But now note that both factors actually belong to $A$ since $b-a$ divides $b^k-a^k$ for all $k\geq0$.