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

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 = – 3ab(a + b) – 3cd(c + d)

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

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

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