Câu 1: 3x² - 6xy + 3y² - 12z²

= 3(x² - 2xy + y² - 4z²)

= 3[(x² -2xy+y²)- (2z)²]

= 3[(x-y)² - (2z)²]

=3(x-y-2z)(x-y +2z)

Câu 2: x² - 6xy - 25z²+ 9y²

= x² - 2x.3y + (3y)² - (5z)²

= (x-3y)² - (5z)²

= (x-3y-5z)(x-3y+5z) 

x² - 2xy - 4z² + y²

= (x² - 2xy+y²) - (2z)²

= (x-y)²- (2z)²

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

x²-2xy-4z²+y²

=(x²-2xy+y²)-(2z)²

=(x-y)²-(2z)²

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

a³ - ay - a²x + xy 

=a²(a-x) - y(a-x)

= (a-x)(a² -y) 

x² - z² + y² - 2xy

= (x² - 2xy + y² )- z²

= (x-y)² - z²

= (x-y-z)(x-y+z) 

Theo đề ra, ta có: \(a+b+c=0\Rightarrow a+b=-c\)

\(a^3+b^3+c^3\)

\(=\left(a+b\right)\left(a^2+b^2-ab\right)+c^3\)

\(=\left(a+b\right)[\left(a+b\right)^2-3ab]+c^3\)

\(=-c\left(c^2-3ab\right)+c^3=-c^3+3ab+c^3=3abc\)

5x² - 10xy + 5y² - 20z²

=5(x² - 2xy + y² - 4z²)

= 5[ (x² - 2xy + y²) - (2z)²]

= 5[(x-y)² - (2z)²]

= 5(x-y-2z)(x-y+2z)

5x²-10xy+5y²-20z²

=5(x²-2xy+y²-4z²)

=5[(x-y)²-(2z)²]

=5(x-y-2z)(x-y+2z)

x² - y² - x-y

= (x-y)(x+y) - (x+y)

=(x+y)(x-y-1) 

x²-y²-2xy+y²

=x²-2xy

=x(x-2y)

x² - y² - 2xy + y²

= (x² - 2xy +y²) - y²

= (x-y)² - y²

= (x-y-y)(x-y+y)

= x(x-2y)

27x³-8y³

=(3x)³-(2y)³

=(3x-2y)(9x²+6xy+4y²)

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

x⁶-y⁶

=(x³-y³)(x³+y³)

=(x-y)(x²+xy+y²)(x+y)(x²-xy+y²)

x^6-y^6=(x^3)^2 - (y^3)^2 = (x^3+y^3)(x^3 - y^3) = (x+y)(x^2-xy+y^2)(x-y)(x^2+xy+y^2)