Given the equation (x + y)³ + 1 , we can assume we have two terms here. These are (x + y)³ and 1. Since both terms are perfect cubes, we can use the sum of cubes formula which is:
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where a = (x+y) and b = 1.
Therefore, the factors of (x + y)³ + 1 is:

The factor of (x + y)³ + 1 is (x + y + 1)(x² + 2xy + y² - x - y +1).