If `x^3 + y^3 + 1 = 3xy,` where` x!= y` determine the value of` x + y + 1. `

Answer

To solve the equation \( x^3 + y^3 + 1 = 3xy \) under the condition that \( x \neq y \), we can use a known algebraic identity.

### Step-by-Step Solution:

1. **Identify the identity**:
We know from algebra that:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
In our case, we can let \( a = x \), \( b = y \), and \( c = 1 \). Thus, we can rewrite the equation as:
\[ x^3 + y^3 + 1^3 - 3xy \cdot 1 = 0 \]
This simplifies to:
\[ x^3 + y^3 + 1 - 3xy = 0 \]

2. **Apply the identity**:
According to the identity, for the equation \( x^3 + y^3 + 1 - 3xy = 0 \) to hold, it must be true that:
\[ x + y + 1 = 0 \]
This is because the product \( (x + y + 1) \) must equal zero for the equation to hold true, given \( x \neq y \).

3. **Solve for \( x + y + 1 \)**:
From the equation \( x + y + 1 = 0 \), we can rearrange it to find:
\[ x + y + 1 = 0 \]

4. **Conclusion**:
Therefore, the value of \( x + y + 1 \) is:
\[ \boxed{0} \]