The factors of `(a-b)^3+(b-c)^3+(c-a)^3` are:

Options

A. `(a+b+c)(a^2+b^2+c^2+ab+bc+ac)`
B. `(a+b+c)(a^2+b^2+c^2-ab-bc-ac)`
C. 3(a-b)(b-c)(c-a)
D. (a-b)(b-c)(c-a)

Answer

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

### Step 1: Identify the variables
Let:
- \(x = a - b\)
- \(y = b - c\)
- \(z = c - a\)

### Step 2: Check if \(x + y + z = 0\)
Now, we need to check if \(x + y + z = 0\):
\[ x + y + z = (a - b) + (b - c) + (c - a) \]
When we simplify this:
\[ = a - b + b - c + c - a = 0 \]
Thus, \(x + y + z = 0\).

### Step 3: Use the identity for cubes
Since \(x + y + z = 0\), we can use the identity:
\[ x^3 + y^3 + z^3 = 3xyz \]
Applying this to our expression:
\[ (a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a) \]

### Step 4: Write the final factorization
Therefore, the factors of the expression \((a-b)^3 + (b-c)^3 + (c-a)^3\) are:
\[ 3(a-b)(b-c)(c-a) \]

### Conclusion
The factors of \((a-b)^3 + (b-c)^3 + (c-a)^3\) are \(3(a-b)(b-c)(c-a)\).

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