(x -y)³  - 1 - 3(x -y)(x - y - 1)

= (x -y)³  - 3(x -y)(x - y - 1) - 1

Đặt x - y = t, khi đó ta có:

    t³  -  3t. (t  - 1) - 1

=   t³  -  3t²  + 3t - 1

=  (t  - 1)³

Thay t = x - y vào (t - 1)³ , ta có:  ( x - y - 1)³

Vậy (x -y)³  - 1 - 3(x -y)(x - y - 1) =  ( x - y - 1)³

Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3

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

Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0

=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3

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

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

x^3 - x + 3x^2y + 3xy^2 + y^3 - y

=x³+y³+3x²y+3xy²-x-y

=(x+y)(x²-xy+y²)+3xy(x+y)-(x+y)

=(x+y)(x²-xy+y²+3xy-1)

=(x+y)(x²+2xy+y²-1)

=(x+y)[(x+y)²-1]

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

x^2 + 5x - 6

=x²-x+6x-6

=x.(x-1)+6.(x-1)

=(x-1)(x+6)

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

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

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

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

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

\(=\left(x-y\right)\left(x+5y\right)\)

dat y^2+y=z cho gon

\(z^2-9z+20=z^2-4z-5z+20=z\left(z-4\right)-5\left(z-4\right)=\left(z-4\right)\left(z-5\right)\)

\(thaylai:\left(y^2+y-4\right)\left(y^2+y-5\right)\)

thay hay thi "k" ko hay thi thoi