If LCM and HCF of two expression are (x + 1) and `(x^(4) - 1)`. If one expression is `(x^(2) - 1)`, then another is

Options

A. `x^(3) - 1`
B. `(x - 1) (x^(2) + 1)`
C. `x^(2) + 1`
D. `(x + 1) (x^(2) + 1)`

Answer

To solve the problem, we need to find the second expression given the first expression, the LCM, and the HCF. Let's denote the expressions as follows:

- Let the first expression be \( E_1 = x^2 - 1 \).
- Let the second expression be \( E_2 \) (which we need to find).
- The LCM of the two expressions is given as \( \text{LCM} = x + 1 \).
- The HCF of the two expressions is given as \( \text{HCF} = x^4 - 1 \).

According to the relationship between LCM, HCF, and the expressions, we have:

\[ E_1 \cdot E_2 = \text{LCM} \cdot \text{HCF} \]

### Step 1: Calculate the product of LCM and HCF

First, we need to calculate the product of the LCM and HCF:

\[ \text{LCM} \cdot \text{HCF} = (x + 1)(x^4 - 1) \]

### Step 2: Expand the product

Now, we will expand \( (x + 1)(x^4 - 1) \):

\[ (x + 1)(x^4 - 1) = x \cdot (x^4 - 1) + 1 \cdot (x^4 - 1) = x^5 - x + x^4 - 1 = x^5 + x^4 - x - 1 \]

### Step 3: Set up the equation

Now we set up the equation using the first expression \( E_1 \):

\[ E_1 \cdot E_2 = (x^2 - 1) \cdot E_2 = x^5 + x^4 - x - 1 \]

### Step 4: Solve for \( E_2 \)

To find \( E_2 \), we can rearrange the equation:

\[ E_2 = \frac{x^5 + x^4 - x - 1}{x^2 - 1} \]

### Step 5: Factor the denominator

Notice that \( x^2 - 1 \) can be factored as \( (x - 1)(x + 1) \).

### Step 6: Simplify the expression

Now, we will simplify \( E_2 \):

1. Factor \( x^5 + x^4 - x - 1 \) if possible.
2. Perform polynomial long division or synthetic division to divide \( x^5 + x^4 - x - 1 \) by \( x^2 - 1 \).

After performing the division, we find:

\[ E_2 = x^3 + x^2 + 1 \]

### Final Answer

Thus, the second expression \( E_2 \) is:

\[ E_2 = x^3 + x^2 + 1 \]