Ta có:\(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
=>\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
Do đó:
\(a^3+b^3+c^3=3ab\)
=>\(a^3+b^3+c^3-3ab=0\)
=>\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3ab=0\)
=>\(\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2-3ab\left(a+b+c\right)\right]=0\)
=>\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
=>\(\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)=0\)
=>\(\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\right]=0\)
=>\(\left(a+b+c\right)\left[\left(a-b\right)^2+\left(a-c\right)^2+\left(b+c\right)^2\right]=0\)
Do a,b,c đôi một khác nhau nên a-b\(e\)0, a-c\(e\)0, b-c\(e\)0
=>\(\left(a-b\right)^2>0;\left(a-c\right)^2>0;\left(b-c\right)^2>0\)
Suy ra: a+b+c=0 +>Điều phải chứng minh
