Cách 2 :

Đặt a + b - 2c = x ; b + c - 2a = y ; c + a - 2b = z

=> x + y + z = a + b - 2c + b + c - 2a + c + a - 2b = 0

Do x + y + z = 0 => x + y = -z => (x+y)^3 = -z^3

Lại có : x^3 + y^3 + z^3

= ( x + y )^3 - 3xy( x + y ) + z^3

= -z^3 - 3xy . ( - z ) + z^3

= 3xyz

hay ( a + b - 2c )^3 + ( b + c - 2a)^3 + ( c + a - 2b)^3

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

Ta có

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

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

Áp dụng vào tổng ta có

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

Đặt

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

=> M=0

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

\(\Rightarrow N=3\left(c-a\right)\left(a-b\right)\left(b-c\right)\)

Để ý : M+N=B

=> \(B=0+3\left(c-a\right)\left(a-b\right)\left(b-c\right)\)

=> \(B=3\left(c-a\right)\left(a-b\right)\left(b-c\right)\)

Phân tích đa thức thành nhân tử

a) x3 +x+2

b) x3-2x-1

c) x3+3x2-4

d) x3+3x2y-9xy2+5y3

a)

\(x^3+x+2\)

\(=\left(x^3+x^2\right)-\left(x^2+x\right)+\left(2x+2\right)\)

\(=x^2\left(x+1\right)-x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x+2\right)\)

b)

\(x^3-2x-1\)

\(=\left(x^3+x^2\right)-\left(x^2+x\right)-\left(x+1\right)\)

\(=x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-1\right)\)

c)

\(x^3-3x^2-4\)

\(=\left(x^3-x^2\right)+\left(4x^2-4x\right)+\left(4x-4\right)\)

\(=x^2\left(x-1\right)+4x\left(x-1\right)+4\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+2.2.x+2^2\right)\)

\(=\left(x-1\right)\left(x+2\right)^2\)

d)

\(x^3-3x^2y-9xy^2+5y^3\)

\(=\left(x^3-x^2y\right)+\left(4x^2y-4xy^2\right)-\left(5xy^2-5y^3\right)\)

\(=x^2\left(x-y\right)+4xy\left(x-y\right)-5y^2\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2-4xy-5y^2\right)\)

\(=\left(x-y\right)^2\left(x-5y\right)\)