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

x3+27+(x+3)(x+9)
=x3+33+(x+3)(x+9)
=(x+3)(x2-3x+9)+(x+3)(x+9)
=(x+3)(x2-2x+18)

\(a)-64+y^2=\left(y-8\right)\left(y+8\right)\)
\(b)-27-x^3=-\left(27+x^3\right)=-\left(3+x\right)\left(9-3x+x^2\right)\)
- 64 + y 2
= - 8 . 8 + y . y
= ( y + 8 ) . ( y - 8 )
-27 - x3
= - ( 27 + x3 )
= - ( 3 + x ) . ( 9 - 3 + x2 )

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

x3-27+3x(x-3)=(x-3)(x2+3x+9)+3x(x-3)
=(x-3)(x2+3x+9+3x)
=(x-3)(x2+6x+9)
=(x-3)(x+3)2

x3 + 27 + (x + 3)(x - 9)
= (x + 3)(x2 + 3x + 9) + (x + 3)(x - 9)
= (x + 3)(x2 + 3x + 9 + x - 9)
= (x + 3)(x2 + 4x)
= x(x + 3)(x + 4)

\(1-27x^3\)
\(=1-\left(3x\right)^3\)
\(=\left(1-3x\right)\left(1+3x+9x^2\right)\)
\(---\)
\(x-3^3+27\)
\(=x-27+27=x\)
\(---\)
\(27x^3+27x^2+9x+1\)
\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)
\(=\left(3x+1\right)^3\)
\(---\)
\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)
\(=\left(\dfrac{x^2}{3}\right)^3-3\cdot\left(\dfrac{x^2}{3}\right)^2\cdot y+3\cdot\dfrac{x^2}{3}\cdot y^2-y^3\)
\(=\left(\dfrac{x^2}{3}-y\right)^3\)
#Ayumu

Ta có : x3 + 27 + (x + 3) (x - 9)
= (x + 3)(x2 - 3x + 9) + (x + 3) (x - 9)
= (x + 3) (x2 - 4x)
= (x + 3) x(x - 4)
Nước kia mình đi sai , cho mình đi lại nha :>
x3 + 27 + ( x + 3 )( x - 9 )
= x3 + 33 + ( x + 3 )( x - 9 )
= ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 9 )
= ( x + 3 )( x2 - 3x + 9 + x - 9 )
= ( x + 3 )( x2 - 2x )
= ( x + 3 )x( x - 2 )
\(x^3+\frac{1}{27}\)
\(=x^3+\left(\frac{1}{3}\right)^3\)
\(=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\left(\frac{1}{3}\right)^2\right)\)
\(=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)
Trả lời hoài mà chẳng ai T bất công thiệt
\(x^3+\frac{1}{27}=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)
(Nếu đúng thì clik cho mình với nhoa!)