Ta có:a2(b3-c3)-a2(b-c)3
=a2(b3-3b2c+3bc2-c3+3b2c-3bc2)-a2(b-c)3
=a2(b-c)3+-a2(b-c)3-3a2bc(b-c)
=3abc(ab-ac)
Tương tự b2(c3-a3)-b2(c-a)3=3abc(bc-ab)
c2(a3-b3)-c2(a-b)3=3abc(ac-bc)
=>a2(b3-c3)-a2(b-c)3+b2(c3-a3)-b2(c-a)3+c2(a3-b3)-c2(a-b)3=3abc(ab-ac+bc-ab+ac-bc)=0
=>a2(b3-c3)+b2(c3-a3)+c2(a3-b3)=a2(b-c)3+b2(c-a)3+c2(a-b)3