Let `alpha ne 1` be a real root of the equation `x^(3) - ax^(2) + ax - 1 = 0,` where a `ne` - 1 is a real number, then a root of this equation, among the following , is :

Options

A. `alpha^2`
B. `-1/(alpha^2)`
C. `1/alpha`
D. `(1)/(alpha^2)`

Answer

To solve the problem, we need to find another root of the equation \( x^3 - ax^2 + ax - 1 = 0 \) given that \( \alpha \) is a root and \( \alpha \neq 1 \).

### Step-by-step Solution:

1. **Substituting the Known Root**: Since \( \alpha \) is a root of the equation, we can substitute \( x = \alpha \) into the equation:
\[ \alpha^3 - a\alpha^2 + a\alpha - 1 = 0 \]

2. **Rearranging the Equation**: Rearranging the equation gives us:
\[ \alpha^3 = a\alpha^2 - a\alpha + 1 \]

3. **Finding Another Root**: We will check if \( \frac{1}{\alpha} \) is another root of the equation. We substitute \( x = \frac{1}{\alpha} \) into the original equation:
\[ \left(\frac{1}{\alpha}\right)^3 - a\left(\frac{1}{\alpha}\right)^2 + a\left(\frac{1}{\alpha}\right) - 1 = 0 \]

4. **Simplifying the Expression**: This simplifies to:
\[ \frac{1}{\alpha^3} - \frac{a}{\alpha^2} + \frac{a}{\alpha} - 1 = 0 \]
Multiplying through by \( \alpha^3 \) to eliminate the denominators gives:
\[ 1 - a\alpha + a\alpha^2 - \alpha^3 = 0 \]

5. **Substituting for \( \alpha^3 \)**: From step 2, we know \( \alpha^3 = a\alpha^2 - a\alpha + 1 \). Substituting this into our equation gives:
\[ 1 - a\alpha + a\alpha^2 - (a\alpha^2 - a\alpha + 1) = 0 \]

6. **Simplifying Further**: This simplifies to:
\[ 1 - a\alpha + a\alpha^2 - a\alpha^2 + a\alpha - 1 = 0 \]
Which simplifies to:
\[ 0 = 0 \]
This confirms that \( \frac{1}{\alpha} \) satisfies the equation.

7. **Conclusion**: Therefore, \( \frac{1}{\alpha} \) is indeed another root of the equation.

### Final Answer:
The root of the equation, among the options given, is \( \frac{1}{\alpha} \).