a: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left\lbrack a^2+2ab+b^2-ac-bc+c^2-3ab\right\rbrack\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
b: Sửa đề: \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Ta có: \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b+c-a\right)\left\lbrack\left(a+b+c\right)^2+a\left(a+b+c\right)+a^2\right\rbrack-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left\lbrack a^2+b^2+c^2+2ab+2ac+2bc+a^2+ab+ac+a^2-b^2+bc-c^2\right\rbrack\)
\(=\left(b+c\right)\left\lbrack3a^2+3ab+3ac+3bc\right\rbrack\)
\(=3\left(b+c\right)\left(a^2+ab+ac+bc\right)\)
\(=3\left(b+c\right)\left\lbrack a\left(a+b\right)+c\left(a+b\right)\right\rbrack\)
=3(b+c)(a+c)(a+b)
=>\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
