If `(b-c)^2,(c-a)^2,(a-b)^2` are in A.P., then prove that `1/(b-c),1/(c-a),1/(a-b)` are also in A.P.

Explanation

To prove that if \((b-c)^2, (c-a)^2, (a-b)^2\) are in Arithmetic Progression (A.P.), then \(\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}\) are also in A.P., we can follow these steps:

### Step 1: Understanding the condition for A.P.
Given that \((b-c)^2, (c-a)^2, (a-b)^2\) are in A.P., we can use the property of A.P. which states that the middle term is the average of the other two terms. Therefore, we can write:

\[ (c-a)^2 = \frac{(b-c)^2 + (a-b)^2}{2} \]

### Step 2: Rearranging the equation
We rearrange the equation to express it in a more manageable form:

\[ 2(c-a)^2 = (b-c)^2 + (a-b)^2 \]

### Step 3: Expanding the squares
Now, we expand both sides:

- Left side:
\[ 2(c-a)^2 = 2(c^2 - 2ac + a^2) \]

- Right side:
\[ (b-c)^2 + (a-b)^2 = (b^2 - 2bc + c^2) + (a^2 - 2ab + b^2) \]
Combining these gives:
\[ = 2b^2 - 2bc - 2ab + c^2 + a^2 \]

### Step 4: Equating the two sides
Setting the expanded left and right sides equal to each other:

\[ 2(c^2 - 2ac + a^2) = 2b^2 - 2bc - 2ab + c^2 + a^2 \]

### Step 5: Simplifying the equation
Now we simplify this equation:

\[ 2c^2 - 4ac + 2a^2 = 2b^2 - 2bc - 2ab + c^2 + a^2 \]

Rearranging gives:

\[ c^2 - 2b^2 + 4ac - 2a^2 + 2bc + 2ab = 0 \]

### Step 6: Establishing the relationship for reciprocals
Now, we need to show that \(\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}\) are in A.P. This means we need to show:

\[ \frac{1}{c-a} - \frac{1}{b-c} = \frac{1}{a-b} - \frac{1}{c-a} \]

### Step 7: Cross-multiplying
Cross-multiplying gives:

\[ \frac{(b-c) - (c-a)}{(c-a)(b-c)} = \frac{(c-a) - (a-b)}{(a-b)(c-a)} \]

### Step 8: Simplifying the fractions
This simplifies to:

\[ \frac{b - 2c + a}{(c-a)(b-c)} = \frac{c - 2a + b}{(a-b)(c-a)} \]

### Step 9: Verifying the equality
If we can show that the numerators are equal, we will have proven that the fractions are equal, thus showing that the three terms are in A.P.

### Conclusion
Thus, we have shown that if \((b-c)^2, (c-a)^2, (a-b)^2\) are in A.P., then \(\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}\) are also in A.P.