Cho a, b, c là những số nguyên thỏa mãn: (1/a+1/b+1/c)^2=1/a^2+1/b^2+!/c^2 

1a+1b+1c2=1a2+1b2+1c2

⇔1a+1b+1c2=1a2+1b2+1c2

Phân tích vế trái ta được: 

1a+1b+1c2=1a+1b2+2ac+2bc+1c2=1a2+2ab+1b2+2ac+2bc+1c2⇒1a2+1b2+1c2+2ab+2ac+2bc=1a2+1b2+1c2⇒2.1ab+1ac+1bc=0⇒cabc+babc+aabc=0⇒a+b+c=0.abc=0⇒a+b=−c⇒−a+b=c

Thay vào ta có: a³ + b³ + c³ = a³ + b³ – (a + b)³ = –3a²b – 3ab² = 3 (–a²b – ab²) ⋮ 3