Question

[Hoàng Phúc]

Phân tích thành nhân tử :

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

Answer 1

[zZz Cool Kid_new zZz]

Đặt \(a+b=m;a-b=n\)

Ta có:\(\Rightarrow\hept{\begin{cases}\left(a+b\right)^2=m^2\\\left(a-b\right)^2=n^2\end{cases}}\Rightarrow\hept{\begin{cases}a^2+2ab+b^2=m^2\\a^2-2ab+b^2=n^2\end{cases}}\Rightarrow\left(a^2+2ab+b^2\right)-\left(a^2-2ab+b^2\right)=m^2-n^2\)

\(\Rightarrow4ab=m^2-n^2\)

Mặt khác :\(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=m\left(n^2+\frac{m^2+n^2}{4}\right)\)

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

\(=\left(m+c\right)^3-4\left[m\left(n^2+\frac{m^2-n^2}{4}\right)+c^3\right]-12abc\)

\(=m^3+3m^2c+3c^2m+c^3-4\left(mn^2+\frac{m^2-n^2}{4}+c^3\right)-12abc\)

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

\(=m^3+3m^2c+3c^2m+c^3-4\cdot\frac{m^3+3mn^2}{4}-4c^3-3cm^2+3cn^2\)

\(=m^3+3cm^2+3c^2m+c^3-m^3-3mn^2-4c^3-3cm^2+3cn^2\)

\(=\left(m^3-m^3\right)+\left(3cm^2-3cm^2\right)+3c^2m+\left(c^3-4c^3\right)+3cn^2-3mn^2\)

\(=3c^2m-3c^3+3cn^2-3mn^2\)

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

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

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

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

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

P/S:Bài giải dài.có j sai thông cảm cho e nha!