Let \[p(x) = \frac{(x + a)^2}{(a - b)(a - c)} + \frac{(x + b)^2}{(b - a)(b - c)} + \frac{(x + c)^2}{(c - a)(c - b)}.\]Then \begin{align*} p(-a) &= \frac{(-a + a)^2}{(a - b)(a - c)} + \frac{(-a + b)^2}{(b - a)(b - c)} + \frac{(-a + c)^2}{(c - a)(c - b)} \\ &= \frac{(b - a)^2}{(b - a)(b - c)} + \frac{(c - a)^2}{(c - a)(c - b)} \\ &= \frac{b - a}{b - c} + \frac{c - a}{c - b} \\ &= \frac{b - a}{b - c} + \frac{a - c}{b - c} \\ &= \frac{b - c}{b - c} \\ &= 1.\end{align*}Similarly, $ p(-b) = p(-c) = 1 $. Since $ p(x) = 1 $ for three distinct values of $ x, $ by the Identity Theorem, $ p(x) = \boxed{1} $ for all $ x $.