a) Đặt a+b-c=x , b+c-a=y, c+a-b=z
\(\Rightarrow\left(a+b+c\right)^3-x^3-y^3-z^3\)
Có x + y +z = a+b-c + b+c-a+c+a-b = a+b+c
\(\Rightarrow\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z^3\right]-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
Áp dụng hằng đẳng thức trên ta có
3(a+b-c+b+c-a)(b+c-a+c+a-b)(a+b-c+c+a-b)
= 3.2b.2c.2a
= 24abc
