Question

Cho a+b+c=0. Tính a³+b³+c³-3ab.

Answer 1

a³ + b³ + c³ - 3ab = [(a + b)³ - 3ab(a + b)] + c³ - 3ab = [(a+ b)³ + c³] - 3ab. (-c) - 3ab

= (a + b+ c)³ - 3(a+ b).c.(a+ b+ c) + 3abc - 3ab = 3abc - 3ab = 3ab.(c - 1)