Ta có :
a3 +b3+c3=3abc
<=> a3+b3+c3 - 3abc =0
Áp dụng (a+b)3-3(a+b) =a3+b3
<=> (a+b)3-3(a+b) +c3-3abc=0
<=> (a+b)3+c3 -3(a+b+c)=0
<=> ( a+b+c) \(\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3\left(a+b+c\right)\)=0
<=> (a+c+b) \(\left[\left(a+b\right)^2-ca-ca+c^2-3\right]=0\)
Vì a+b+c=0 => a3+b3+c3=3abc