What is the value of the expression?
`((a-b)^(3)+(b-c)^(3)+(c-a)^(3))/(3(a-b)(b-c)(c-a))=?`
(a)4
(b)1
(c)2
(d)0

Options

A. 4
B. 1
C. 2
D. 0

Answer

To solve the expression \(\frac{(a-b)^3 + (b-c)^3 + (c-a)^3}{3(a-b)(b-c)(c-a)}\), we can use the identity that states if \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\).

### Step-by-step Solution:

1. **Identify the Terms**:
We have three terms in the numerator: \((a-b)^3\), \((b-c)^3\), and \((c-a)^3\).

2. **Set Up the Identity**:
Notice that if we let:
- \(x = a - b\)
- \(y = b - c\)
- \(z = c - a\)

Then we can see that:
\[ x + y + z = (a-b) + (b-c) + (c-a) = 0 \]

3. **Apply the Identity**:
Since \(x + y + z = 0\), we can apply the identity:
\[ x^3 + y^3 + z^3 = 3xyz \]
Therefore, we can rewrite the numerator:
\[ (a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a) \]

4. **Substitute Back into the Expression**:
Substitute this result back into the original expression:
\[ \frac{(a-b)^3 + (b-c)^3 + (c-a)^3}{3(a-b)(b-c)(c-a)} = \frac{3(a-b)(b-c)(c-a)}{3(a-b)(b-c)(c-a)} \]

5. **Simplify**:
The \(3(a-b)(b-c)(c-a)\) in the numerator and denominator cancel out:
\[ = 1 \]

### Final Answer:
Thus, the value of the expression is \(1\).