See this problem very similar .

It can be easily shown that dist(x,y)=dist(x,z)=dist(y,z) is possible only when there exists node v, such that dist(x,v)=dist(y,v)=dist(z,v).
We iterate v from 1 to n and for each we want to know the amount of good x,y,z triples.
Let's delete v from the tree, then it will break down into several trees, let's denote them as T1,T2..Tk.
We will run DFS for each tree T1,T2..Tk and write down pairs (depth; which_tree).
Now we want to calculate the number of triples q,w,e such that q.depth=w.depth=e.depth, and q.which_tree,w.which_tree,e.which_tree are all different.
Let's work with each depth separately.
Considering depth D, we will write down "which_tree" values for all pairs which have .depth=D.
Now, let's merge same which_tree values into one, remembering the amount.
Now we're only left with the following problem: given A1,A2,A3...Az, calculate the sum of A[i]*A[j]*A[k] for all possible different i,j,k.
After calculating this value for each v, just add it to the answer.
And don't forget to return v back to your tree. =)
Probably my code will help you to understand better: link