Introduction:
To prove that (b-c), (c-a), and (a-b) are in arithmetic progression (A.P.), we will use the given information that (b-c)^2, (c-a)^2, and (a-b)^2 are in A.P. We will break down the proof into the following sections:

Section 1: Deriving the Arithmetic Progression:
1. Let (b-c)^2 be the first term of the A.P., (c-a)^2 be the second term, and (a-b)^2 be the third term.
2. The common difference of the A.P. can be represented as d.
3. Using the formula for an A.P., the second term is given by the first term plus the common difference: (c-a)^2 = (b-c)^2 + d.
4. Similarly, the third term is given by the second term plus the common difference: (a-b)^2 = (c-a)^2 + d.

Section 2: Simplifying the Equations:
1. Expanding the squares in the above equations, we get:
- (c^2 - 2bc + b^2) = (b^2 - 2bc + c^2) + d
- (a^2 - 2ac + c^2) = (c^2 - 2ac + a^2) + d
2. Simplifying the above equations, we obtain:
- -2bc = d
- -2ac = d

Section 3: Proving Arithmetic Progression:
1. Subtracting the second equation from the first equation, we get:
-2bc - (-2ac) = d - d
=> 2ac - 2bc = 0
=> ac - bc = 0
=> c(a-b) = 0
2. Since c ≠ 0 (as it is a non-zero term in (b-c)^2), we can divide both sides by c:
a - b = 0
=> a = b
3. Therefore, if a = b, the common difference d becomes:
d = -2bc
4. Substituting the value of d in the equations of the A.P., we find:
(c-a)^2 = (b-c)^2 - 2bc
(a-b)^2 = (c-a)^2 - 2ca
5. Simplifying the above equations, we get:
(c-a)^2 = (b-c)^2 - 2bc
(a-b)^2 = (c-a)^2 - 2ca
=> (c-a)^2 - (b-c)^2 = 2bc
=> (a-b)^2 - (c-a)^2 = 2ca
6. The above equations indicate that (c-a)^2 - (b-c)^2 and (a-b)^2 - (c-a)^2 are equal to 2bc and 2ca, respectively.
=> (c-a)^2 - (b-c)^2 = (a-b)^2 - (c-a)^2
=> (c-a)^2 - (a-b)^2 = (b-c)^2 - (c-a)^2