`(a-b)^3+(b-c)^3+(c-a)^3`

Answer

To solve the expression \((a-b)^3 + (b-c)^3 + (c-a)^3\), we can use a known algebraic identity. Here’s a step-by-step solution:

### Step 1: Recognize the Identity
We can use the identity for the sum of cubes:
\[ x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz) \]
In our case, let:
- \(x = a - b\)
- \(y = b - c\)
- \(z = c - a\)

### Step 2: Calculate \(x + y + z\)
Now, we calculate:
\[ x + y + z = (a - b) + (b - c) + (c - a) \]
Notice that all terms cancel out:
\[ x + y + z = a - b + b - c + c - a = 0 \]

### Step 3: Apply the Identity
Since \(x + y + z = 0\), we can simplify the expression:
\[ x^3 + y^3 + z^3 = 3xyz \]
Thus, we need to find \(xyz\):
\[ xyz = (a-b)(b-c)(c-a) \]

### Step 4: Substitute Back
Now substituting back, we have:
\[ (a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a) \]

### Final Answer
The final result is:
\[ (a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a) \]