a, Ta có : \(a^3+b^3+c^3-3abc\)
= \(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
= \(\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3ab\left(a+b+c\right)\)
= \(\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right)\)