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(a+b+c)^3 thì viết được thành [(a+b)+c)]^3 rồi AD hằng đẳng thức để tính. Còn với (a^3+b^3+c^3) ta viết được (a+b)^3 -3a^2b -3ab^2 + c^3=(a+b)^3 -3ab(a+b)+c^3 ...thay vào rồi đổi biến
Đặt A là tên biểu thức; \(a+b-c=x;b+c-a=y;c+a-b=z\)
Khi đó \(x+y+z=a+b-c+b+c-a+c+a-b=a+b+c\)
=>\(A=\left(x+y+z\right)^3-x^3-y^3-z^3=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\)
\(=3.2b.2c.2a=24abc\)
A = ( a + b + c )3 + ( a - b - c )3 + ( b - c - a )3 + ( c - a - b )3
= [ ( a + b ) + c ]3 + [ ( a - b ) - c ]3 + [ ( - c ) - ( a - b ) ] 3 + [ c - ( a + b ) ]3
= ( a + b )3 + 3.( a + b )2.c + 3.( a + b ).c2 + c3 + ( a - b )3 - 3.( a - b )2.c + 3.( a - b ).c2 - c3 + ( - c3 ) + 3.( a - b )2.c - 3.( a - b ).c2 -(a- b)3
+ c3 + 3.( a + b )2.c - 3.( a + b ).c2 - ( a + b )3
= 6.( a + b )2 .c
\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\))
\(=a^3\left(b-c\right)-b^3\left(a-c\right)+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left[\left(b-c\right)+\left(a-b\right)\right]+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left(b-c\right)-b^3\left(a-b\right)+c^3\left(a-b\right)\)
\(=\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(a^2+ab+b^2\right)-\left(b^2+bc+c^2\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2-c^2+ab-bc\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(a-c\right)\left(a+c\right)+b\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left[a-b+b-c\right]+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left(a-b\right)-b^3\left(b-c\right)+c^3\left(a-b\right)\)
\(=\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab+b^2-b^2-bc-c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab-bc-c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(a-c\right)\left(a+c\right)+b\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
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!