Solve in $\mathbb{R}$ the equation :
$(x+1)^5 + (x+1)^4(x-1) + (x+1)^3(x-1)^2 +$ $ (x+1)^2(x-1)^3 + (x+1)(x-1)^4 + (x-1)^5 =$ $ 0$.
1. We start with the given equation:
   \[
   (x+1)^5 + (x+1)^4(x-1) + (x+1)^3(x-1)^2 + (x+1)^2(x-1)^3 + (x+1)(x-1)^4 + (x-1)^5 = 0
   \]

2. Notice that the given expression can be rewritten using the binomial theorem. Specifically, we can factor out the common terms:
   \[
   \sum_{k=0}^{5} (x+1)^{5-k} (x-1)^k
   \]

3. This sum is a telescoping series, which can be simplified using the identity for the difference of powers:
   \[
   \sum_{k=0}^{5} (x+1)^{5-k} (x-1)^k = \frac{(x+1)^6 - (x-1)^6}{(x+1) - (x-1)}
   \]

4. Simplify the denominator:
   \[
   (x+1) - (x-1) = 2
   \]

5. Therefore, the equation becomes:
   \[
   \frac{(x+1)^6 - (x-1)^6}{2} = 0
   \]

6. For the fraction to be zero, the numerator must be zero:
   \[
   (x+1)^6 - (x-1)^6 = 0
   \]

7. This implies:
   \[
   (x+1)^6 = (x-1)^6
   \]

8. Since both sides are raised to an even power, we consider the absolute values:
   \[
   |x+1| = |x-1|
   \]

9. This gives us two cases to consider:
   - Case 1: \( x+1 = x-1 \)
   - Case 2: \( x+1 = -(x-1) \)

10. Solving Case 1:
    \[
    x+1 = x-1 \implies 1 = -1 \quad \text{(which is impossible)}
    \]

11. Solving Case 2:
    \[
    x+1 = -(x-1) \implies x+1 = -x+1 \implies 2x = 0 \implies x = 0
    \]

12. Therefore, the only solution is:
    \[
    x = 0
    \]

The final answer is \(\boxed{x = 0}\)