If `(a^(2)-1)x^(2)+(a-1)x+(a^(2)-4a+3)=0` is an identity in x, then the value of a is/are

Options

A. `-1`
B. 3
C. 1
D. `-1,1,3`

Answer

To solve the equation \((a^{2}-1)x^{2}+(a-1)x+(a^{2}-4a+3)=0\) being an identity in \(x\), we need to ensure that the coefficients of \(x^2\), \(x\), and the constant term are all equal to zero.

### Step-by-Step Solution:

1. **Identify the coefficients**:
- The coefficient of \(x^2\) is \(a^2 - 1\).
- The coefficient of \(x\) is \(a - 1\).
- The constant term is \(a^2 - 4a + 3\).

2. **Set the coefficients to zero**:
Since the equation is an identity in \(x\), we set each coefficient to zero:
\[ a^2 - 1 = 0 \quad (1) \]
\[ a - 1 = 0 \quad (2) \]
\[ a^2 - 4a + 3 = 0 \quad (3) \]

3. **Solve equation (1)**:
\[ a^2 - 1 = 0 \implies a^2 = 1 \implies a = 1 \text{ or } a = -1 \]

4. **Solve equation (2)**:
\[ a - 1 = 0 \implies a = 1 \]

5. **Solve equation (3)**:
\[ a^2 - 4a + 3 = 0 \]
Factoring gives:
\[ (a - 1)(a - 3) = 0 \implies a = 1 \text{ or } a = 3 \]

6. **Combine the solutions**:
From the three equations, we have:
- From (1): \(a = 1\) or \(a = -1\)
- From (2): \(a = 1\)
- From (3): \(a = 1\) or \(a = 3\)

The common solution across all equations is \(a = 1\).

### Final Answer:
The value of \(a\) that satisfies the identity is:
\[ \boxed{1} \]