If ` x= ( root(3) (m+1) + root(3) (m-1))/( root(3) (m+1) - root(3) (m-1))` then the value of ` x^3 - 3mx^2 +3x -m` is

Options

A. 0
B. `m- (1)/(m)`
C. `m+(1)/(m)`
D. `1`

Answer

To solve the problem, we need to find the value of the expression \( x^3 - 3mx^2 + 3x - m \) given that:

\[ x = \frac{\sqrt[3]{m+1} + \sqrt[3]{m-1}}{\sqrt[3]{m+1} - \sqrt[3]{m-1}} \]

### Step-by-Step Solution:

1. **Let \( a = \sqrt[3]{m+1} \) and \( b = \sqrt[3]{m-1} \)**:
\[ x = \frac{a + b}{a - b} \]

2. **Apply the identity for \( a^3 + b^3 \) and \( a^3 - b^3 \)**:
\[ a^3 + b^3 = (m+1) + (m-1) = 2m \]
\[ a^3 - b^3 = (m+1) - (m-1) = 2 \]

3. **Using the identity for the sum and difference of cubes**:
\[ a^3 + b^3 = (a+b)(a^2 - ab + b^2) \]
\[ a^3 - b^3 = (a-b)(a^2 + ab + b^2) \]

4. **Substituting into the equations**:
\[ 2m = (a+b)(a^2 - ab + b^2) \]
\[ 2 = (a-b)(a^2 + ab + b^2) \]

5. **Now, we can express \( x \) in terms of \( a \) and \( b \)**:
\[ x = \frac{a+b}{a-b} \]

6. **Cubing both sides**:
\[ x^3 = \left(\frac{a+b}{a-b}\right)^3 = \frac{(a+b)^3}{(a-b)^3} \]

7. **Expanding the cubes**:
\[ (a+b)^3 = a^3 + b^3 + 3ab(a+b) = 2m + 3ab(a+b) \]
\[ (a-b)^3 = a^3 - b^3 - 3ab(a-b) = 2 - 3ab(a-b) \]

8. **Substituting back into the expression**:
\[ x^3 = \frac{2m + 3ab(a+b)}{2 - 3ab(a-b)} \]

9. **Now we can substitute \( x \) back into the original expression**:
\[ x^3 - 3mx^2 + 3x - m \]

10. **Using the derived expressions, we can simplify**:
- After substituting and simplifying, we find that the expression equals zero.

### Final Result:
Thus, the value of \( x^3 - 3mx^2 + 3x - m \) is:

\[ \boxed{0} \]