If `a^(2)-4a+1=4`, then the value of `(a^(3)-a^(2)+a-1)/(a^(2)-1)(a^(2)ne1)`

Explanation

To solve the equation \( a^2 - 4a + 1 = 4 \) and find the value of \( \frac{a^3 - a^2 + a - 1}{(a^2 - 1)(a^2 \neq 1)} \), we will follow these steps:

### Step 1: Simplify the given equation
Start with the equation:
\[ a^2 - 4a + 1 = 4 \]
Subtract 4 from both sides:
\[ a^2 - 4a + 1 - 4 = 0 \implies a^2 - 4a - 3 = 0 \]

### Step 2: Factor the quadratic equation
Now, we will factor the quadratic equation:
\[ a^2 - 4a - 3 = 0 \]
We look for two numbers that multiply to \(-3\) and add to \(-4\). The factors are \(-5\) and \(1\):
\[ (a - 5)(a + 1) = 0 \]
Thus, we have:
\[ a - 5 = 0 \quad \text{or} \quad a + 1 = 0 \]
This gives us the solutions:
\[ a = 5 \quad \text{or} \quad a = -1 \]

### Step 3: Calculate the expression for both values of \( a \)
We need to evaluate \( \frac{a^3 - a^2 + a - 1}{(a^2 - 1)} \).

#### For \( a = 5 \):
1. Calculate \( a^3 - a^2 + a - 1 \):
\[ 5^3 - 5^2 + 5 - 1 = 125 - 25 + 5 - 1 = 104 \]
2. Calculate \( a^2 - 1 \):
\[ 5^2 - 1 = 25 - 1 = 24 \]
3. Now substitute into the expression:
\[ \frac{104}{24} = \frac{13}{3} \]

#### For \( a = -1 \):
1. Calculate \( a^3 - a^2 + a - 1 \):
\[ (-1)^3 - (-1)^2 + (-1) - 1 = -1 - 1 - 1 - 1 = -4 \]
2. Calculate \( a^2 - 1 \):
\[ (-1)^2 - 1 = 1 - 1 = 0 \]
Since \( a^2 \neq 1 \) is a condition, we cannot use \( a = -1 \) in our final answer.

### Step 4: Final answer
Thus, the only valid solution is when \( a = 5 \):
\[ \text{The value of } \frac{a^3 - a^2 + a - 1}{(a^2 - 1)} = \frac{13}{3} \]

### Summary
The final answer is:
\[ \frac{13}{3} \]