帮我做几道因式分解题目（一题10分,共6题）

1.y(y+1)(x²+1）+x(2y²+2y+1)+1
=y(y+1)(x^2+1)+2x(y+1)y+x+1
=y(y+1)(x+1)^2+x+1
=(x+1)(xy^2+xy+y^2+y+1)
2.6x²-5xy-6y²-2xz-23yz-20z²
=6x^2-(5y+2z)x-6y^2-23yz-20z^2
=6[x^2-(5/6y+1/3z)x+(3/2y+2z)(-2/3y-5/3)z
=6(x-3/2y+2z)(x+2/3y+5/3z)
=(2x-3y+4z)(3x+2y+5z)
3.2x²-7xy+6y²+2x-y-12
=(x-2y)(2x-3y)+2x-y-12
=(x-2y)(2x-3y)+6x-4x-9y+8y-12
=(x-2y)(2x-3y)+6x-9y-4x+8y-12
=(x-2y)(2x-3y)+3(2x-3y)-4(x-2y+3)
=(x-2y+3)(2x-3y)-4(x-2y+3)
=(x-2y+3)(2x-3y-4)
4.（x²）²-x²y²+（4y²）²
=(x^2)^2+8x^2y^2+(4y^2)^2-9x^2y^2
=(x^2+4y^2)^2-(3xy)^2
=(x^2+3xy+4y^2)(x^2-3xy+4y^2)
5.（x²）²+5x²+9
=(x^2)^2+6x^2+9-x^2
=(x^2+3)^2-x^2
=(x^2+x+3)(x^2-x+3)
6.x³+6x²+11x+6
=x^3+x^2+5x^2+5x+6x+6
=x^2(x+1)+5x(x+1)+6(x+1)
=(x^2+5x+6)(x+1)
=(x+1)(x+2)(x+3)