Define $ P(\tau ) = (\tau + 1)^3$ . If $x + y = 0$, what is the minimum possible value of $P(x) + P(y)$?
1. Given the function \( P(\tau) = (\tau + 1)^3 \), we need to find the minimum possible value of \( P(x) + P(y) \) given that \( x + y = 0 \).

2. Since \( x + y = 0 \), we can substitute \( y = -x \).

3. Therefore, we need to evaluate \( P(x) + P(-x) \):
   \[
   P(x) + P(-x) = (x + 1)^3 + (-x + 1)^3
   \]

4. Expand both terms using the binomial theorem:
   \[
   (x + 1)^3 = x^3 + 3x^2 + 3x + 1
   \]
   \[
   (-x + 1)^3 = (-x)^3 + 3(-x)^2 + 3(-x) + 1 = -x^3 + 3x^2 - 3x + 1
   \]

5. Add the expanded forms:
   \[
   P(x) + P(-x) = (x^3 + 3x^2 + 3x + 1) + (-x^3 + 3x^2 - 3x + 1)
   \]

6. Combine like terms:
   \[
   P(x) + P(-x) = x^3 - x^3 + 3x^2 + 3x^2 + 3x - 3x + 1 + 1
   \]
   \[
   P(x) + P(-x) = 6x^2 + 2
   \]

7. The expression \( 6x^2 + 2 \) is a quadratic function in terms of \( x \). The minimum value of a quadratic function \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). Here, \( a = 6 \) and \( b = 0 \), so the vertex is at:
   \[
   x = -\frac{0}{2 \cdot 6} = 0
   \]

8. Substitute \( x = 0 \) into the quadratic expression to find the minimum value:
   \[
   6(0)^2 + 2 = 2
   \]

9. Therefore, the minimum possible value of \( P(x) + P(y) \) is \( 2 \).

The final answer is \( \boxed{2} \).