Integral closure of $A \subseteq B$ is a subring of $B$

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So I have the following exercise:

Let $A \subseteq B$ both rings, prove that $\hat{A} = \left\{\alpha \in B: f(\alpha) = 0 \right\}$ is a subring of $B$, with $f(x) \in A[x]$ some monic polynomial.

And I have to prove it with the clasic subring test (substraction and multiplication closure in $\hat{A}$) and I already got the multiplication closure but still stuck with the substraction closure:

$$f(\alpha - \beta) = (\alpha - \beta)^n + a_{n-1}(\alpha - \beta)^{n-1} + \dots + a_1(\alpha - \beta) + a_0, \ \alpha, \beta \in \hat{A}$$

I thought about using binomial newton but I was wondering if there was another easier way to do it.