Question

Phân tích đa thức thành thân tử:

a. \(A=a^3-b^3-c^3-3abc\)

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

c. \(C=x^3+3x^2-4\)

Answer 1

a)  \(A=a^3-b^3-c^3-3abc\)

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

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

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

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

Answer 2

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

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

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

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

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

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

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

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

Answer 3

c)  \(C=x^3+3x^2-4\)

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

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

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

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

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

Answer 4

câu a của mk là từ HĐT:

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

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

để từ đó suy ra: 

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

sau đó mk thay vào biểu thức ban đầu