Ta có: \(A=\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)

Áp dụng hằng đẳng thức: \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)

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

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

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

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

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

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

Phân tích đa thức thành nhân tử

A) x^2 +3x -16

B) 15x^2 -x-6

C) x^2-4x+13

a, \(x^2+3x-16\)

\(=x^2+1,5x+1,5x+2,25-18,25\)

\(=\left(x+1,5\right)^2-\left(\sqrt{18,25}\right)^2\)

\(=\left(x+1,5+\sqrt{18,25}\right)\left(x+1,5-\sqrt{18,25}\right)\)

b, \(15x^2-x-6\)

\(=15x^2-10x+9x-6\)

\(=5x\left(3x-2\right)+3\left(3x-2\right)\)

\(=\left(3x-2\right)\left(5x+3\right)\)

c, \(x^2-4x+13\)

Pt này bạn xem lại nha!

Phân tích các đa thức sau thành nhân tử ( đặt biến phụ )

a. \(\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)

b. \(\left(x^2+x\right)^2+4x^2+4x-12\)

c. \(x^4+2x^3+5x^2+4x-12\)

d.\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

e. \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

f. \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

a) \(\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)

Đặt \(x^2+x=y\) ta được:

\(y^2-14y+24\)

\(=x\left(y-12\right)-2\left(y-12\right)\)

\(=\left(y-2\right)\left(y-12\right)\)

Thay ngược trở lại:

\(\left(x^2+x-2\right)\left(x^2+x-12\right)\)

\(=\left(x-1\right)\left(x+2\right)\left(x-3\right)\left(x+4\right)\)

d) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+10\right)+1\)

Đặt \(x^2+5x+4=a\) được:

\(a\left(a+6\right)+1\)

\(=a^2+6a+1\)

\(=a^2+2.a.3+3^2-8\)

\(=\left(a+3\right)^2-\left(\sqrt{8}\right)^2\)

\(=\left(a+3-\sqrt{8}\right)\left(a+3+\sqrt{8}\right)\)

Mấy câu kia tương tự.