Đặt `a+b=x, b+c=y, c+a=z`
`->x+y+z=2 (a+b+c)`
`(a+b)^3 +(b+c)^3 + (c+a)^3 - 8 (a+b+c)^3`
`= x^3 + y^3 + z^3 - 2^3 (a+b+c)^3`
`=x^3 +y^3 +z^3 - [2 (a+b+c)]^3`
`=x^3 +y^3+z^3 - (x+y+z)^3`
`= x^3 + y^3 +z^3 - [x^3 +y^3 +z^3 + 3 (x+y) (y+z) (x+z)]`
`= -3 (x+y)(y+z)(x+z)`
`= -3 (2b + a+c) (2c+a+b) (2a +b+c)`
Đặt : \(\hept{\begin{cases}a+b=x\\b+c=y\\c+a=z\end{cases}}\Rightarrow\left(a+b\right)^3+\left(b+c\right)^3+\left(a+c\right)^3-8\left(a+b+c\right)^3=x^3+y^3+z^3-\left(x+y+z\right)^3\)
\(=x^3+y^3+z^3-\left(x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\right)=-3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(=-3\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)\)
