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

\(\Rightarrow\)\(a^3\left(b-c\right)+bc\left(b^2-c^2\right)-a\left(b^3-c^3\right)\)

\(\Rightarrow\)\(a^3\left(b-c\right)+bc\left(b-c\right)\left(b+c\right)-a\left(b-c\right)\left(b^2+bc+c^2\right)\)

\(\Rightarrow\)\(\left(b-c\right)\left(a^3+bc\left(b+c\right)-a\left(b^2+bc+c^2\right)\right)\)

\(\Rightarrow\)\(\left(b-c\right)\left(a^3+b^2c+bc^2-ab^2-abc-ac^2\right)\)

\(\Rightarrow\)\(\left(b-c\right)\left(bc\left(c-a\right)+b^2\left(c-a\right)-a\left(c^2-a^2\right)\right)\)

\(\Rightarrow\)\(\left(b-c\right)\left(c-a\right)\left(bc+b^2-a\left(c+a\right)\right)\)

\(\Rightarrow\)\(\left(b-c\right)\left(c-a\right)\left(bc+b^2-ac-a^2\right)\)

\(\left(b-c\right)\left(c-a\right)\left(b^2-a^2+c\left(b-a\right)\right)=\left(b-c\right)\left(c-a\right)\left(b-a\right)\left(a+b+c\right)\)

\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)

\(=\left(a+b-2c+b+c-2a\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]+\left(c+a-2b\right)^3\)

\(=\left(c+a-2b\right)^3-\left(a-2b+c\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(c+a-2b\right)^2-\left(a+b-2c\right)^2+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(c+a-2b+a+b-2c\right)\left(c+a-2b-a-b+2c\right)+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b\right)-\left(a+b-2c\right)\left(2a-b-c\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b-a-b+2c\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(5c-a-4b\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(b+c-2a\right)\left(a+4b-5c\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(a+4b-5c-b-c+2a\right)\)

\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(3a+3b-6c\right)\)

\(=3\left(c+a-2b\right)\left(b+c-2a\right)\left(a+b-2c\right)\)

\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)

Đặt: \(a+b-2c=x;b+c-2a=y;c+a-2b=z\)

\(\Rightarrow B=x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

Ta thấy: \(x+y+z=a+b-2c+b+c-2a+c+a-2b=0\)

\(x+y=a+b-2c+b+c-2a=2b-a-c\)

\(y+z=b+c-2a+c+a-2b=2c-a-b\)

\(z+x=c+a-2b+a+b-2c=2a-b-c\)

Thay vào B \(\Rightarrow B=0-3\left(2b-a-c\right)\left(2c-a-b\right)\left(2a-b-c\right)\)

Vậy \(B=-3\left(2b-a-a\right)\left(2c-a-b\right)\left(2a-b-c\right).\)

Đặt A là tên biểu thức; \(a+b-c=x;b+c-a=y;c+a-b=z\)

Khi đó \(x+y+z=a+b-c+b+c-a+c+a-b=a+b+c\)

=>\(A=\left(x+y+z\right)^3-x^3-y^3-z^3=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)

\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)

\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)

\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

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

\(=3.2b.2c.2a=24abc\)

1) \(\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(c+b+a\right)\)

đặt a+b-c=x;b+c=y;c+a-b=z có x+y+z=a+b+c

có:(x+y+z)^3-x^3-y^3-z^3=3(x+y)(y+z)(x+z)

tách đoạn đầu đi mình đánh mỏi tay lắm :v

=24abc nhé :v