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 

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

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

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

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

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

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

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

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

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

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

mình lớp 6 ko biết làm

a(b-c)³+b(c-a)³+c(a-b)³

=a(b-c)³-b[(a-b)+(b-c)]+c(a-b)³

=a(b-c)³-b[(a-b)³+3(a-b)²(b-c)+3(a-b)(b-c)²+(b-c)³]

                                  +c(a-b)³

=a(b-c)³-b(a-b)³+3b(a-b)²(b-c)+3b(a-b)(b-c)²+b(b-c)³

                                  +c(a-b)³

=(b-c)³(a-b)-(a-b)³(b-c)-3b(a-b)(b-c)(a-b+b-c)

=(b-c)³(a-b)-(a-b)³(b-c)-3b(a-b)(b-c)(a-c)

=(a-b)(b-c)[(b-c)²-(a-b)²-3b(a-c)]

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

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

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

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

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

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

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

(a+b+c)3−a3−b3−c3(a+b+c)3−a3−b3−c3

=a3+3a2(b+c)+3a(b+c)2+(b+c)3−a3−b3−c3=a3+3a2(b+c)+3a(b+c)2+(b+c)3−a3−b3−c3

=3(b+c)(a2+ab+ac)+b3+3b2c+3bc2+c3−b3−c3=3(b+c)(a2+ab+ac)+b3+3b2c+3bc2+c3−b3−c3

=3(b+c)(a2+ab+ac+bc)=3(b+c)(a2+ab+ac+bc)

=3(b+c)[a(a+b)+c(a+b)]=3(b+c)[a(a+b)+c(a+b)]

=3(b+c)(a+b)(a+c)

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