Belarusian Mathematical Olympiad

Set $A = a + b + c$. Then
$$\begin{aligned} M &= \frac{(a+b-c)^2}{(a-c)(b-c)} + \frac{(b+c-a)^2}{(b-a)(c-a)} + \frac{(c+a-b)^2}{(c-b)(a-b)} = \\ &= \frac{(A-2c)^2}{(a-c)(b-c)} + \frac{(A-2a)^2}{(b-a)(c-a)} + \frac{(A-2b)^2}{(c-b)(a-b)} = \\ &= \frac{(A-2c)^2(b-a) + (A-2a)^2(c-b) + (A-2b)^2(a-c)}{(a-b)(b-c)(c-a)} = \frac{L}{(a-b)(b-c)(c-a)} \end{aligned}$$

We have
$$\begin{aligned} L &= (A^2 - 4Ac + 4c^2)(b-a) + (A^2 - 4Aa + 4a^2)(c-b) + \\ &\quad (A^2 - 4Ab + 4b^2)(a-c) = A^2((b-a) + (c-b) + (a-c)) - \\ &\quad -2A(c(b-a) + a(c-b) + b(a-c)) + 4(c^2(b-a) + a^2(c-b) + b^2(a-c)) = \\ &= -4A(cb - ca + ac - ab + ba - bc) + 4N = 4N. \end{aligned}$$

Further
$$\begin{aligned} N &= c^2(b-a) + a^2(c-b) + b^2(a-c) = c^2(b-a) + (a^2c - a^2b + b^2a - b^2c) = \\ &= c^2(b-a) + (a^2c - b^2c) - (a^2b - b^2a) = c^2(b-a) + c(a+b)(a-b) - ab(a-b) = \\ &= (a-b)(ca + cb - ab - c^2) = (a-b)(a(c-b) + c(b-c)) = (a-b)(b-c)(c-a). \end{aligned}$$
Therefore, $M = 4$.