a^3-2a^2b+ab^2-b+1=a^3-a^2b-a^2b+ab^2-b+1
=(a^3+1)-a^2b-b+ab^2-a^2b=(a^3+1)-(a^2b+b)+(ab^2-a^2b)
=(a+1)(a^2-a+1)-b(a^2+1)-ab(a-b)
=(a+1)[(a^2+1)-a]-b(a^2+1)-ab(a-b)
=(a+1)(a^2+1)-a(a+1)-b(a^2+1)-ab(a-b)
=(a+1)(a^2+1)-b(a^2+1)-a(a+1))-ab(a-b)
=(a^2+1)(a-b+1)-a^2-a-a^2b+ab^2
=(a^2+1)(a-b+1)-(a^2+a^2b)+(ab^2-a)
=(a^2+1)(a-b+1)-a^2(1+b)+a(b+1)(b-1)
=(a^2+1)(a-b+1)-a(b+1)(a-b+1)
=(a-b+1)(a^2-ab-a+1)
mx^2+23x-35
令mx^2+23x-35=（x+7)(mx+b）(b為待定系數(shù))
（x+7)(mx+b）=mx^2+(7m+b)x+7b
對(duì)應(yīng)系數(shù)：得7m+b=23,7b=-35,聯(lián)立解得：b=-5,m=4;
故：m=4.