(1)x^6+64y^6+12x^2y^2-1

题目

(1)x^6+64y^6+12x^2y^2-1
(2)4m^2-14mx+49nx-14mn
(3)a^2b^2c-ab^3c+a^2bc^2-ab^2c^2
(4)x^5-x^3-x^2-1
(5)1+a+b+c+ab+bc+ac+abc
(6)ab(c^2-d^2)-cd(a^2-b^2)
(7)(x^2+1)^2-x^2+x(x-2)(x^2+x+1)

答案

(1)x^6+64y^6+12x^2y^2-1
=x^6+64y^6+12x^4y^2+12x^2y^4-12x^4y^2-12x^2y^4+12x^2y^2-1
=(x^2+y^2)^3-1-12x^2y^2(x^2+y^2-1)
=(x^2+y^2-1)[(x^2+y^2)^2+x^2+y^2+1]-12x^2y^2(x^2+y^2-1)
=(x^2+y^2-1)[x^4+y^4+x^2+y^2-10x^2y^2+1]
=(x^2+y^2-1)[x^4+y^4-10x^2y^2+x^2+y^2+1]
(2)4m^2-14mx+49nx-14mn
=4m^2-14m(x+n)+49nx
=(2m-7n)(2m-7x)
(3)a^2b^2c-ab^3c+a^2bc^2-ab^2c^2
=abc(ab-b^2+ac-bc)
=abc[a(b+c)-b(b+c)]
=abc(b+c)(a-b)
(4)x^5-x^3-x^2+1
=x^2(x^3-1)-(x^3-1)
=(x+1)(x-1)(x-1)(x^2+x+1)
(5)1+a+b+c+ab+bc+ac+abc
=1+c+a+b+ab(1+c)+c(a+b)
=(1+c)(1+ab)+(a+b)(1+c)
=(1+c)(1+ab+a+b)
=(1+c)[(1+b)+a(1+b)]
=(1+c)(1+b)(1+a)
(7)(x^2+1)^2-x^2+x(x-2)(x^2+x+1)
=(x^2+1+x)(x^2+1-x)+x(x-2)(x^2+x+1)
=(x^2+x+1)(2x^2-3x+1)
=(x^2+x+1)(2x-1)(x-1)