Answer: \(0\)
Solution:
I originally tried solving this with imaginary numbers with the first equation, then gave up because I'm pretty bad at dealing with imaginary numbers, then thought about it a little bit, and thought of this answer.
a¹⁸⁸¹ + a¹⁸⁸² + a¹⁸⁸³ = a¹⁸⁸³ + a¹⁸⁸² + a¹⁸⁸¹ = a¹⁸⁸¹(a² + a + 1) = a¹⁸⁸¹ x 0 = 0
You can do the same thing twice again, except for the second time make it a to the power of 1884, 1885, and 1886, and for the third make it a to the power of 1887, 1888, and 1889.
All three give 0, so 0 + 0 + 0 = \(0\)