\(-x^3+9x^2-27x+27=\left(3-x\right)^3\)

\(-x^3+9x^2-27x+27\)

\(=-x^3+3x^2+6x^2-18x-9x+27\)

\(=-x^2\left(x-3\right)+6x\left(x-3\right)-9\left(x-3\right)\)

\(=-\left(x-3\right)\left(x^2-6x+9\right)\)

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

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

1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)

2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)

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

3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)

1) Ta có: \(x^3+2x^2-6x-27\)

\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

2: Ta có: \(9x^2+6x-4y^2-4y\)

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

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

Câu c đề sai thì phải bạn ạ.

a) \(27+x^3=\left(x+3\right)\left(x^2-3x+9\right)\)

b) \(-x^3+12x^2-48x+64=\left(4-x\right)^3\)

c) \(27+27x+9x^2=9\left(x^2+3x+3\right)\)

\(27+27x+9x^2+x^3\)

\(=x^3+9x^2+27x+27\)

\(=x^3+3x^2+6x^2+18x+9x+27\)

\(=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2+6x+9\right)\)

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

Nếu nhìn kĩ thì bạn sẽ thấy đây là hằng đẳng thức nhé !

\(x^3+9x^2+27x+27=x^3+3.3.x^2+3.3^2.x+3^3\)

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

= (x+3)³

27+27x+9x²+x³

= x³+27x+9x²+27

=(x+3)³

CHÚC BẠN HỌC TỐT

a) \(x^2-xz-9y^2+3yz\)

\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)

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

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

c) \(x^3+2x^2-6x-27\)

\(=\left(x^3-27\right)+\left(2x^2-6x\right)\)

\(=\left(x-3\right)\left(x^2-3x+9\right)+2x\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-3x+9+2x\right)\)

\(=\left(x-3\right)\left(x^2-x+9\right)\)

b) \(x^3-x^2-5x+125\)

\(=\left(x^3+125\right)-\left(x^2+5x\right)\)

\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(5+x\right)\)

\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

\(8-27x^3\)

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

\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)

a) \(8-27x^3=\left(2-x\right)\left(4+6x+9x^2\right)\)

b) \(27+27x+9x^2+x^3=\left(3+x\right)^3\)

c) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

27x³ +\(\frac{y^{3^{ }}}{8}\)

1/1\27 là jz?