Giải quyết bằng toán này bằng cách đặt ẩn phụ. 

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Đặt  \(a+b=m\)   \(;\) \(a-b=n\)  thì  \(4ab=\left(a^2+2ab+b^2\right)-\left(a^2-2ab+b^2\right)=\left(a+b\right)^2-\left(a-b\right)^2\) , tức là  \(4ab=m^2-n^2\) và \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(a+b\right)\left[\left(a^2-2ab+b^2\right)+ab\right]=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\) ,

tức là  \(a^3+b^3=m\left(n^2+\frac{m^2-n^2}{4}\right)\)

Ta có:

\(A=\left(a+b+c\right)^3-4\left(a^3+b^3+c^3\right)-12abc\)

\(=\left(m+c\right)^3-4\left[m\left(n^2+\frac{m^2-n^2}{4}\right)+c^3\right]-3c\left(m^2-n^2\right)\)

\(=m^3+3m^2c+3mc^2+c^3-4mn^2-m^3+mn^2-4c^3-3m^2c+3n^2c\)

\(=3mc^2-3c^3-3mn^2+3n^2c\)

\(=3\left(mc^2-c^3-mn^2+n^2c\right)\)

\(=3\left[c^2\left(m-c\right)-n^2\left(m-c\right)\right]\)

\(=3\left(m-c\right)\left(c^2-n^2\right)=3\left(m-c\right)\left(c-n\right)\left(c+n\right)\)

Do đó,   \(A=3\left(a+b-c\right)\left(c-a+b\right)\left(c+a-b\right)\)