Cho a3 + b3 + c3 = 3abc. Tính B = b=(1 a/b)(1 b/c)(1 c/a) .

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:

a+b=−ca+c=−bb+c=−a

B = 1+ab1+bc1+ca=a+bb.b+ca.a+cb=−c−a−babc=−1

Vậy B = – 1.