Hoặc
Ta có:
P=abc2+bca2+acb2P=abcc3+abca3+abcb3=abc1c3+1a3+1b3
Vì 1a+1b+1c=0
⇔1a+1b=−1c⇔1a+1b3=−1c3⇔1a3+1b3+3ab1a+1b=−1c3⇔1a3+1b3+1c3+3ab−1c=0⇔1a3+1b3+1c3−3abc=0
⇔1a3+1b3+1c3=3abc (1)
Thay (1) vào P ta được: P=abc.3abc=3