1.将原式+X-X,构成X^4-X+2001X^2+2001X+2001=X(X^3-1)+2001(X^2+X+1)=X(X-1)(X^2+X+1)+2001(X^2+X+1)=(X^2+X+1)(X^2-X+2001)
2.a(a+b)^2-(a+b)^2 +1
=(ab-1)(a+b)^2+1
=(ab-1)a^2+(ab-1)b^2+2ab(ab-1)+1
=(ab-1)a^2+(ab-1)b^2+2(ab)^2-2ab+1
=(ab-1)a^2+(ab-1)b^2+(ab)^2+(ab)^2-2ab+1
=(ab-1)a^2+(ab-1)b^2+(ab)^2+(ab^2-2ab+1)
=(ab-1)a^2+(ab-1)b^2+(ab)^2+(ab-1)^2
=[(ab-1)a^2+(ab)^2]+[(ab-1)b^2+(ab-1)^2]
=a^2[(ab-1)+b^2]+(ab-1)*[(ab-1)+b^2]
=[a^2+(ab-1)]*[b^2+(ab-1)]
3.x^2+xy-6y^2+x+13y-6
x^2+xy-6y=[x+3y][x-2y]
然后设原式=[x+3y+m][x-2y+n]
=[x+3y][x-2y]+n[x+3y]+m[x-2y]+mn
=[x+3y][x-2y]+[n+m]x+[3n-2m]y+mn
则有：m+n=1 3n-2m=13 mn=-6
m=-2
n=3
所以原式=[x+3y-2][x-2y+3]