\(=6x\left(-y^2+x^2+2x+1\right)\\ =6x\left[\left(x^2+2x+1\right)-y^2\right]\\ =6x\left[\left(x+1\right)^2-y^2\right]\\ =6x\left(x+1-y\right)\left(x+1+y\right)\)

\(=\left(x^3-6x^2+12x-8\right)+1\\ =\left(x-2\right)^3+1\\ =\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)\\ =\left(x-1\right)\left(x^2-5x+7\right)\)

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

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

\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

=(x^3+6x^2+12x+8)+y^3

=(x^3+3x^2+3x2^2+2^3)+y^3

=(x+2)^3+y^3

=(x+2+y)((x+2)^2-(x+2)y+y^2)

=(x+2+y)(x^2+4x+4-xy-2y+y^2)

=(x+2+y)(x^2+y^2-xy+4x-2y+4)

ko bít đâu nha : lớp 5

x³ - 6x² + 12x - 8

= x³ - 2x² - 4x² + 4x + 8x - 8

= (x³ - 2x²) - (4x² - 8x) + (4x - 8)

= x².(x - 2) + 4x.(x - 2) + 4.(x - 2)

= (x - 2).(x² + 4x + 4)

= (x - 2).(x² + 2x + 2x + 4)

= (x - 2).[x.(x + 2) + 2.(x + 2)]

= (x - 2).(x + 2).(x + 2)

= (x - 2).(x + 2)²

những bạn nhầm đề bài r bạn

x^3+y^3+6x^2+12x+8

=(x^3+6x^2+12x+8)+y^3

=(x+2)^3+y^3

=(x+2+y).(x^2+4x+4+y^2-xy-2y)

x³ - 6x² + 12x - 8

= x³ - 2x² - 4x² + 4x + 8x - 8

= (x³ - 2x²) - (4x² - 8x) + (4x - 8)

= x².(x - 2) + 4x.(x - 2) + 4.(x - 2)

= (x - 2).(x² + 4x + 4)

= (x - 2).(x² + 2x + 2x + 4)

= (x - 2).[x.(x + 2) + 2.(x + 2)]

= (x - 2).(x + 2).(x + 2)

= (x - 2).(x + 2)²

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

a: \(x^2+12x+36=0\) 

=>\(x^2+2\cdot x\cdot6+6^2=0\)

=>\(\left(x+6\right)^2=0\)

=>x+6=0

=>x=-6

b: \(4x^2-4x+1=0\)

=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

=>\(\left(2x-1\right)^2=0\)

=>2x-1=0

=>2x=1

=>x=1/2

c: \(x^3+6x^2+12x+8=0\)

=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)

=>\(\left(x+2\right)^3=0\)

=>x+2=0

=>x=-2