HCF and LCM of two algebraic expressions are respectively (a + 1) and `(a^(3) +a^(2)-a-1)`. If one of the expression is `a^(2)-1`, then what is the second expression?

Options

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

Answer

To solve the problem, we need to find the second algebraic expression given the HCF and LCM of two expressions and one of the expressions.

### Given:
- HCF = \( a + 1 \)
- LCM = \( a^3 + a^2 - a - 1 \)
- One expression = \( a^2 - 1 \)

### Step 1: Use the relationship between HCF, LCM, and the two expressions.
The relationship states that:
\[ \text{HCF} \times \text{LCM} = \text{First Expression} \times \text{Second Expression} \]
Let the second expression be \( E_2 \). Therefore, we can write:
\[ (a + 1)(a^3 + a^2 - a - 1) = (a^2 - 1) \times E_2 \]

### Step 2: Substitute the known values.
Substituting the known values into the equation gives us:
\[ (a + 1)(a^3 + a^2 - a - 1) = (a^2 - 1) \times E_2 \]

### Step 3: Simplify the left-hand side.
First, we need to simplify \( a^2 - 1 \):
\[ a^2 - 1 = (a + 1)(a - 1) \]
Now, substituting this back into the equation:
\[ (a + 1)(a^3 + a^2 - a - 1) = (a + 1)(a - 1) \times E_2 \]

### Step 4: Cancel \( (a + 1) \) from both sides.
Assuming \( a + 1 \neq 0 \), we can cancel \( a + 1 \) from both sides:
\[ a^3 + a^2 - a - 1 = (a - 1) \times E_2 \]

### Step 5: Solve for \( E_2 \).
Now, we can express \( E_2 \) as:
\[ E_2 = \frac{a^3 + a^2 - a - 1}{a - 1} \]

### Step 6: Perform polynomial long division.
Now we will perform polynomial long division of \( a^3 + a^2 - a - 1 \) by \( a - 1 \).

1. Divide the leading term: \( a^3 \div a = a^2 \).
2. Multiply \( a^2 \) by \( a - 1 \): \( a^3 - a^2 \).
3. Subtract:
\[ (a^3 + a^2) - (a^3 - a^2) = 2a^2 \]
4. Bring down the next term: \( 2a^2 - a = 2a^2 - a \).
5. Divide the leading term: \( 2a^2 \div a = 2a \).
6. Multiply \( 2a \) by \( a - 1 \): \( 2a^2 - 2a \).
7. Subtract:
\[ (2a^2 - a) - (2a^2 - 2a) = a \]
8. Bring down the next term: \( a - 1 \).
9. Divide the leading term: \( a \div a = 1 \).
10. Multiply \( 1 \) by \( a - 1 \): \( a - 1 \).
11. Subtract:
\[ (a - 1) - (a - 1) = 0 \]

Thus, we have:
\[ E_2 = a^2 + 2a + 1 \]

### Step 7: Recognize the result.
The expression \( a^2 + 2a + 1 \) can be factored as:
\[ E_2 = (a + 1)^2 \]

### Final Answer:
The second expression is:
\[ \boxed{(a + 1)^2} \]