Question

Phân tích đa thức thành nhân tử x^3-3x^2y+3xy^2-y^3-z^z^3                         
x^2-y^2+8x+6y+7

 

 

 

 

Answer 1

x³ - 3x²y + 3xy² - y³ - z³

= (x³ - 3x²y + 3xy² - y³) - z³

= (x - y)³ - z³

= (x - y - z)[(x - y)² + (x - y)z + z²]

= (x - y - z)(x² - 2xy + y² + xz - yz + z³)

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x² - y² + 8x + 6y + 7

= (x² + 8x + 16) - (y² - 6y + 9)

= (x + 4)² - (y - 3)²

= (x + 4 - y + 3)(x + 4 + y - 3)

= (x - y + 7)(x + y + 1)

Answer 2

a: \(=\left(x^3-3x^2y+3xy^2-y^3\right)-z^3\)

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

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

\(=\left(x-y-z\right)\left(x^2-2xy+y^2+xz-yz+z^2\right)\)

b: \(=x^2+8x+16-y^2+6y-9\)

=(x+4)^2-(y-3)^2

=(x+4+y-3)(x+4-y+3)

=(x+y+1)(x-y+7)