đặt a-b=x

    b-c=y

    c-a=z

x+y+z=0 => x+y=-z <=> x^3 + y^3 +3xy(x+y) =-z^3 <=> x^3 +y^3 +z^3 =3xyz ( vì x+y=-z)

thế vào pt B = 3(a-b)(b-c)(c-a) 

k mình nha đúng nhất nè :)))))))))))))))

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

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

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

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

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

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

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

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

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

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

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 

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