The number of integral values which can be taken by the expression, `f (x) = (x ^(3)-1)/((x-1) (x ^(2) -x+1))` for `x in R,` is:
1
2
3
infinite

Options

A. A) 1
B. B) 2
C. C) 3
D. D) infinite

Answer

To solve the problem, we need to analyze the function given by:

\[ f(x) = \frac{x^3 - 1}{(x - 1)(x^2 - x + 1)} \]

### Step 1: Simplify the Expression

First, we can factor the numerator \(x^3 - 1\):

\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]

So we can rewrite \(f(x)\) as:

\[ f(x) = \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x^2 - x + 1)} \]

### Step 2: Cancel the Common Factor

Since \(x - 1\) is a common factor in the numerator and denominator (for \(x \neq 1\)), we can cancel it out:

\[ f(x) = \frac{x^2 + x + 1}{x^2 - x + 1} \quad \text{for } x \neq 1 \]

### Step 3: Analyze the Simplified Function

Now we need to analyze the function:

\[ f(x) = \frac{x^2 + x + 1}{x^2 - x + 1} \]

### Step 4: Finding Integral Values

To find the integral values of \(f(x)\), we can set \(f(x) = k\) where \(k\) is an integer:

\[ \frac{x^2 + x + 1}{x^2 - x + 1} = k \]

Cross-multiplying gives:

\[ x^2 + x + 1 = k(x^2 - x + 1) \]

Rearranging this leads to:

\[ (1 - k)x^2 + (1 + k)x + (1 - k) = 0 \]

### Step 5: Determine the Discriminant

For \(f(x)\) to have integral values, the discriminant of this quadratic equation must be a perfect square:

\[ D = (1 + k)^2 - 4(1 - k)(1 - k) \]

Calculating the discriminant:

\[ D = (1 + k)^2 - 4(1 - k)^2 \]

Expanding both terms:

\[ D = 1 + 2k + k^2 - 4(1 - 2k + k^2) \]
\[ D = 1 + 2k + k^2 - 4 + 8k - 4k^2 \]
\[ D = -3k^2 + 10k - 3 \]

### Step 6: Set Discriminant as a Perfect Square

We need to find integer values of \(k\) such that \(-3k^2 + 10k - 3\) is a perfect square.

### Step 7: Testing Integer Values

We can test integer values for \(k\):

1. For \(k = 1\):
\[ D = -3(1)^2 + 10(1) - 3 = 4 \quad (\text{Perfect square}) \]

2. For \(k = 2\):
\[ D = -3(2)^2 + 10(2) - 3 = 7 \quad (\text{Not a perfect square}) \]

3. For \(k = 3\):
\[ D = -3(3)^2 + 10(3) - 3 = 12 \quad (\text{Not a perfect square}) \]

4. For \(k = 4\):
\[ D = -3(4)^2 + 10(4) - 3 = 19 \quad (\text{Not a perfect square}) \]

5. For \(k = 5\):
\[ D = -3(5)^2 + 10(5) - 3 = 28 \quad (\text{Not a perfect square}) \]

Continuing this process, we find that there are more integer values that can be tested, leading to the conclusion that there are infinite integral values for \(f(x)\).

### Conclusion

Thus, the number of integral values that can be taken by the expression \(f(x)\) is:

\[ \text{Infinite} \]