Ta có: \(\left(a+b+c\right)^3\)
\(=\left(a+b\right)^3+3\cdot\left(a+b\right)^2\cdot c+3\cdot\left(a+b\right)\cdot c^2+c^3\)
\(=\left(a+b\right)^3+3c\left(a^2+2ab+b^2\right)+3c^2\left(a+b\right)+c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3a^2c+6abc+3b^2c+3ac^2+3bc^2+c^3\)
\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2\) +6abc
Ta có: \(\left(a+b+c\right)^3-\left(a+b\right)^3-\left(b+c\right)^3-\left(a+c\right)^3\)
\(=\left(a+b+c\right)^3-\left(a^3+3a^2b+3ab^2+b^3\right)-\left(b^3+3b^2c+3bc^2+c^3\right)-\left(a^3+3a^2c+3ac^2+c^3\right)\)
\(=\left(a+b+c\right)^3-2a^3-2b^3-2c^3-3a^2b-3ab^2-3a^2c-3ac^2-3b^2c-3bc^2\)
\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2\) +6abc-\(2a^3-2b^3-2c^3-3a^2b-3ab^2-3a^2c-3ac^2-3b^2c-3bc^2\)
\(=6abc-a^3-b^3-c^3\)
