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\(x^3-3x^2+3x-1-y^3\\ =\left(x-1\right)^3-y^3=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\\ =\left(x-y-1\right)\left(x^2-2x+1+xy-y+y^2\right)\)

\(x^3+3x^2+3x+1-y^3\\ =\left(x+1\right)^3-y^3=\left(x+1-y\right)\left[\left(x+1\right)^2+y\left(x+1\right)+y^2\right]\\ =\left(x-y+1\right)\left(x^2+2x+1+xy+y+y^2\right)\)

a) x³-3x²+3x-1-y³    =(x-1)³ - y³

= (x-1)³ - 3.(x-1)².y + 3.(x-1). y² - y

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

x3 + 3x2 + 3x + 1 = x3 + 3x2.1 + 3x.12 + 13 = (x + 1)3

d) x3 + 3x2 – 3x – 1

= (x3 - 1) + (3x2 - 3x)

= (x - 1)(x2 + x + z) + 3x(x - 1)

= (x - 1)(x2 + 4x + 1)

a) x3 + 3x2 – 3x – 9

= (x3 + 3x2) - (3x + 9)

= x2(x + 3) - 3(x + 3)

= (x + 3)(x2 - 3)

= (x + 3)(x + √3)(x - √3)

a) \(A=x^2-6x+9-9y^2\)

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

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

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

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

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

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

a, \(A=\left(x-3\right)^2-9y^2=\left(x-3-3y\right)\left(x-3+3y\right)\)

b, \(B=\left(x-1\right)^3+2\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left[\left(x-1\right)^2+2\left(x+1\right)\right]\)

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

a) x² - 9

= x² - 3²

= (x - 3)(x + 3)

b) 4x² - 1

= (2x)² - 1²

= (2x - 1)(2x + 1)

c) x⁴ - 16

= (x²)² - 4²

= (x² - 4)(x² + 4)

= (x² - 2²)(x² + 4)

= (x - 2)(x + 2)(x + 4)

d) x² - 4x + 4

= x² - 2.x.2 + 2²

= (x - 2)²

e) x³ - 8

= x³ - 2³

= (x - 2)(x² + 2x + 4)

f) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³