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

Options

A. a) `-1`
B. b) `1`
C. c) `3`
D. d) All of the above

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 given quadratic equation can be rewritten in the standard form \(Ax^2 + Bx + C = 0\), where:
- \(A = a^2 - 1\)
- \(B = a - 1\)
- \(C = a^2 - 4a + 3\)

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

3. **Solve equation (1)**:
From \(a^2 - 1 = 0\):
\[ a^2 = 1 \]
Taking the square root of both sides, we get:
\[ a = 1 \quad \text{or} \quad a = -1 \]

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

5. **Solve equation (3)**:
From \(a^2 - 4a + 3 = 0\):
We can factor this quadratic:
\[ (a - 3)(a - 1) = 0 \]
Therefore, we have:
\[ a = 3 \quad \text{or} \quad a = 1 \]

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

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

### Conclusion:
Thus, the value of \(a\) that satisfies the given equation as an identity in \(x\) is:
\[ \boxed{1} \]