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