Let's construct a polynomial $$f(x)=\frac{(x+a)(x+b)}{(c-a)(c-b)}+\frac{(x+b)(x+c)}{(a-b)(a-c)}+\frac{(x+c)(x+a)}{(b-c)(b-a)}-1$$ Therefore, the to-be-proved claim is equivalent to showing $f(x)=0$ is an identity. Because $f(x)$ is a quadratic polynomial with respect to $x$, therefore it is sufficient to show that there exist three distinct zero points. \begin{align} f(-a) &=\frac{(-a+a)(-a+b)}{(c-a)(c-b)}+\frac{(-a+b)(-a+c)}{(a-b)(a-c)}+\frac{(-a+c)(-a+a)}{(b-c)(b-a)}-1\\ &=\frac{(b-a)(c-a)}{(a-b)(a-c)}-1\\ &=0 \end{align} By symmetry, we can assert $f(-b)=f(-c)=0$ must hold too. It is clear that $a$, $b$, and $c$ are distinct because otherwise the given expression is not defined. Therefore, we can conclude $f(x)=0$ always holds.