Give the function `f(x)=(x+1) (x^(3)-1)`. Find `f(-1),f(a+1), f(a)+1`.

Answer

To solve the problem, we need to evaluate the function \( f(x) = (x + 1)(x^3 - 1) \) at three different points: \( f(-1) \), \( f(a + 1) \), and \( f(a) + 1 \).

### Step 1: Calculate \( f(-1) \)

1. Substitute \( x = -1 \) into the function:
\[ f(-1) = (-1 + 1)((-1)^3 - 1) \]
2. Simplify:
\[ f(-1) = (0)(-1 - 1) = 0 \cdot (-2) = 0 \]

### Step 2: Calculate \( f(a + 1) \)

1. Substitute \( x = a + 1 \) into the function:
\[ f(a + 1) = ((a + 1) + 1)((a + 1)^3 - 1) \]
This simplifies to:
\[ f(a + 1) = (a + 2)((a + 1)^3 - 1) \]
2. Now calculate \( (a + 1)^3 \):
\[ (a + 1)^3 = a^3 + 3a^2 + 3a + 1 \]
3. Substitute this back into the function:
\[ f(a + 1) = (a + 2)(a^3 + 3a^2 + 3a + 1 - 1) \]
This simplifies to:
\[ f(a + 1) = (a + 2)(a^3 + 3a^2 + 3a) \]

### Step 3: Calculate \( f(a) + 1 \)

1. Substitute \( x = a \) into the function:
\[ f(a) = (a + 1)(a^3 - 1) \]
2. Now add 1:
\[ f(a) + 1 = (a + 1)(a^3 - 1) + 1 \]

### Summary of Results

- \( f(-1) = 0 \)
- \( f(a + 1) = (a + 2)(a^3 + 3a^2 + 3a) \)
- \( f(a) + 1 = (a + 1)(a^3 - 1) + 1 \)