Cho a + b + c + d = 0. Chứng minh rằng a3 + b3 + c3 + d3 = 3(c + d)(ab – cd).

Ta có a + b + c + d = 0

⇔ a + b = – c – d

⇔ (a + b)³ = (– c – d)³

⇔ a³ + b³ + 3ab(a + b) = – c³ – d³ – 3cd(c + d)

⇔a³ + b³ + c³ + d³ = – 3cd(c + d) – 3ab(a + b)

⇔a³ + b³ + c³ + d³ = – 3cd(c + d) + 3ab(c + d)

⇔a³ + b³ + c³ + d³ = 3(c + d)(ab – cd)

Vậy a³ + b³ + c³ + d³ = 3(c + d)(ab – cd).