Explanation:
let's do the squaring on the right side :
1/(b-c)² + 1/(b-c)(c-a) + 1/(b-c)(a-b) + 1/(b-c)(c-a) + 1/(c-a)² +
+ 1/(c-a)(a-b) + 1/(b-c)(a-b) + 1/(c-a)(a-b) + 1/(a-b)² =
= 1/(b-c)² + 1/(c-a)² + 1/(a-b)² +
+ 2/(b-c)(c-a) + 2/(b-c)(a-b) + 2/(c-a)(a-b)
that means, that in order for the original equation to be true
2/(b-c)(c-a) + 2/(b-c)(a-b) + 2/(c-a)(a-b) = 0
as the first part of the right side is exactly the left side of the original equation.
when
2/(b-c)(c-a) + 2/(b-c)(a-b) + 2/(c-a)(a-b) = 0
then also
1/(b-c)(c-a) + 1/(b-c)(a-b) + 1/(c-a)(a-b) = 0
since we can divide on both sides, we can also multiply and keep the overall equation true.
so, we multiply both sides by e.g. (b-c)(c-a) and get
1/1 + (b-c)(c-a)/(b-c)(a-b) + (b-c)(c-a)/(c-a)(a-b) = 0
1/1 + (c-a)/(a-b) + (b-c)/(a-b) = 0
let's now multiply with (a-b) :
(a-b) + (c-a) + (b-c) = 0
a-a + b-b + c-c = 0
0 = 0
therefore, the original expression is true.