=(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)

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

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

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

=\(x^3+3.2.x^2+3.2^2.x+2^3+y^3\)

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

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

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

a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

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

c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)

đưa về dạng hằng đẳng thức thứ 4 lập phương của 1 tổng

\(x^3+6x^2+12x+8\)

\(=\left(x+2\right)^3\)
 

\(x^3+6x^2+12x+8=x^3+2.3x^2+2.3^2x+2^3=\left(x+2\right)^3\)

xong ròi k1 mình nha bn thanks

x^3 + 3.x^2.2 + 3.x.2^2 + 2^3

= (x+2)^3

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