Question

[Lê Tài Bảo Châu]

\(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)

\(=a^3+3a^2b+3ab^2+b^3+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)

\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)

\(=3\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc\right)\)

\(=3\left[\left(a^2b+ab^2\right)+\left(a^2c+abc\right)+\left(ac^2+bc^2\right)+\left(b^2c+abc\right)\right]\)

\(=3\left[ab\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)+bc\left(a+b\right)\right]\)

\(=3\left(a+b\right)\left(ab+ac+c^2+bc\right)\)

\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+b\right)\)

Answer 1

[๖ۣۜRᶤℵ﹏❖(๖ۣۜBảo)]

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