What is the common zero of the polynomial

`x^(3) + 1, x^(2) - 1` and `x^(2) + 2x + 1?`

Answer

To find the common zero of the polynomials \( x^3 + 1 \), \( x^2 - 1 \), and \( x^2 + 2x + 1 \), we will follow these steps:

### Step 1: Find the zeros of each polynomial.

1. **Polynomial 1: \( x^3 + 1 \)**
- We can factor this polynomial using the sum of cubes formula:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
- Here, \( a = x \) and \( b = 1 \):
\[ x^3 + 1 = (x + 1)(x^2 - x + 1) \]
- Setting \( x + 1 = 0 \) gives us:
\[ x = -1 \]
- The quadratic \( x^2 - x + 1 \) does not have real roots (discriminant \( (-1)^2 - 4 \cdot 1 \cdot 1 = -3 < 0 \)).

2. **Polynomial 2: \( x^2 - 1 \)**
- This can be factored using the difference of squares:
\[ x^2 - 1 = (x - 1)(x + 1) \]
- Setting \( x - 1 = 0 \) gives \( x = 1 \).
- Setting \( x + 1 = 0 \) gives \( x = -1 \).

3. **Polynomial 3: \( x^2 + 2x + 1 \)**
- This can be factored as a perfect square:
\[ x^2 + 2x + 1 = (x + 1)^2 \]
- Setting \( (x + 1)^2 = 0 \) gives:
\[ x + 1 = 0 \Rightarrow x = -1 \]

### Step 2: Identify the common zero.

From our findings:
- The first polynomial \( x^3 + 1 \) has a zero at \( x = -1 \).
- The second polynomial \( x^2 - 1 \) has zeros at \( x = -1 \) and \( x = 1 \).
- The third polynomial \( x^2 + 2x + 1 \) has a zero at \( x = -1 \) (with multiplicity 2).

The common zero among all three polynomials is:
\[ \boxed{-1} \]