First, we rearrange the terms:
x³ - y³ + x - y = 1948
Now factor the difference of cubes:(x - y)(x² + xy + y²) + x - y = 1948
(x - y)(x² + xy + y² + 1) = 1948
Since x and y are both integers, we can conclude that both factors in the above equation must be integers as well. Thus we need to find two numbers whose product is 1948:
1 x 1948 = 1948
2 x 974 = 1948
4 x 487 = 1984
Clearly (x - y) will be the smaller factor, so we need to look for possibilities where
x = y + 1
x = y + 2
x = y + 4

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If x = y + 1, then
(y + 1)² + (y + 1)y + y² + 1 = 1948
y² + 2y + 1 + y² + y + y² + 1 = 1948
3y² + 3y - 1946 = 0
Clearly there is no integer value of y for which this is true, since 1946 is not divisible by 3.

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If x = y + 2, then
(y + 2)² + (y + 2)y + y² + 1 = 974
y² + 4y + 4 + y² + 2y + y² + 1 = 974
3y² + 6y - 969 = 0
y² + 2y - 323 = 0
This gives us y = 17, and therefore x = 19.

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If x = y + 4, then
(y + 4)² + (y + 4)y + y² + 1 = 487
y² + 8y + 16 + y² + 4y + y² + 1 = 487
3y² + 12y - 470 = 0

As with the first possibility, there are no integer values for which this is true.

Thus, the only solution is: (19, 17)