If `a+b+c=3x`, then find the value of `(x - a)^3+ (x - b)^3 +(x-c)^3`

Options

A. 0
B. `3(x-a) (x-b) (x-c)`
C. 1
D. `(x-a) (x-b) (x-c)`

Answer

To solve the problem, we need to find the value of \((x - a)^3 + (x - b)^3 + (x - c)^3\) given that \(a + b + c = 3x\).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ a + b + c = 3x \]

2. **Rewrite the expression we need to evaluate**:
We want to find:
\[ (x - a)^3 + (x - b)^3 + (x - c)^3 \]

3. **Use the identity for the sum of cubes**:
The identity for the sum of cubes states:
\[ p^3 + q^3 + r^3 - 3pqr = (p + q + r)(p^2 + q^2 + r^2 - pq - qr - rp) \]
Here, let \(p = x - a\), \(q = x - b\), and \(r = x - c\).

4. **Calculate \(p + q + r\)**:
\[ p + q + r = (x - a) + (x - b) + (x - c) = 3x - (a + b + c) = 3x - 3x = 0 \]

5. **Substituting into the identity**:
Since \(p + q + r = 0\), we can simplify the identity:
\[ (x - a)^3 + (x - b)^3 + (x - c)^3 = 3(x - a)(x - b)(x - c) \]

6. **Finding \((x - a)(x - b)(x - c)\)**:
We can express this product as:
\[ (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc \]
Since \(a + b + c = 3x\), we have:
\[ (x - a)(x - b)(x - c) = x^3 - 3x^2x + (ab + ac + bc)x - abc = x^3 - 3x^3 + (ab + ac + bc)x - abc \]
This simplifies to:
\[ -2x^3 + (ab + ac + bc)x - abc \]

7. **Final expression**:
Therefore, substituting back, we have:
\[ (x - a)^3 + (x - b)^3 + (x - c)^3 = 3(-2x^3 + (ab + ac + bc)x - abc) \]

### Conclusion:
The final value of \((x - a)^3 + (x - b)^3 + (x - c)^3\) is:
\[ 3(-2x^3 + (ab + ac + bc)x - abc) \]