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1 Answer

w³-29=(w-∛29)(w²+w∛29+(∛29)²) based on w³-a³=(w-a)(w²+aw+a²) where a=∛29. (∛29)² can be written 29^⅔.

Further factorisation is possible but this involves complex numbers:

w²+aw+a²/4-a²/4+a²=(w+½a)²+3a²/4=(w+½a+½ai√3)(w+½a-½ai√3).

w³-29=(w-∛29)(w+½∛29+½i∛29√3)(w+½∛29-½i∛29√3).

by Top Rated User (774k points)

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