Cho a3 + b3 + c3 = 3abc. Tính B = b=(1 a/b)(1 b/c)(1 c/a) .
Cho a3 + b3 + c3 = 3abc. Tính B = .
Ta có:
a3 + b3 + c3 = 3abc
⇔ (a + b)3 – 3ab(a + b) + c3 – abc = 0
⇔ (a + b + c)[(a + b)2 + (a + b)c + c2] – 3ab(a + b + c) = 0
⇔ (a + b + c) [(a + b)2 + (a + b)c + c2 – 3abc] = 0
Do (a + b)2 + (a + b)c + c2 – 3abc > 0 nên a + b + c = 0
Suy ra:
B =
Vậy B = – 1.