Question

请大家解两道数学题，先化简再求值，写出详细过程

（x^2-xy+y^2)^2(x^2+xy+y^2)^2(x-y)^2(x+y)^2,其中x=2,y=1

(a+b)(a^4+b^4)(a-b)(a^2+b^2)其中a=1,b=-2

Answer 1

1、因为x=2,y=1,所以有
（x^2-xy+y^2)^2(x^2+xy+y^2)^2(x-y)^2(x+y)^2
=[(x+y)(x^2-xy+y^2)]^2[(x-y)(x^2+xy+y^2)]^2
=(x^3+y^3)^2(x^3-y^3)^2
=(x^6-y^6)^2
=(2^6-1^6)^2
=3969
2、因为a=1,b=-2,所以有
(a+b)(a^4+b^4)(a-b)(a^2+b^2)
=(a^4+b^4)(a^2+b^2)(a+b)(a-b)
=(a^4+b^4)(a^2+b^2)(a^2-b^2)
=(a^4+b^4)(a^4-b^4)
=a^8-b^8
=1^8-(-2)^8
=-255

Answer 2

（x^2-xy+y^2)^2(x^2+xy+y^2)^2(x-y)^2(x+y)^2
=[(x+y)^2x^2-xy+y^2)^2][(x-y)^2(x^2+xy+y^2)^2]
=(x^3+y^3)^2(x^3-y^3)^2
=(x^6-y^6)^2
=(2^6-1^6)^2
=63^2
=3969

(a+b)(a^4+b^4)(a-b)(a^2+b^2)
=(a+b)(a-b)(a^2+b^2)(a^4+b^4)
=(a^2-b^2)(a^2+b^2)(a^4+b^4)
=a^8-b^8
=1^8-(-2)^8
=1-256
=-255

Answer 3

一、二楼已解，本人放弃，不好意思。

Answer 4

(x^2 - xy + y^2)^2 (x^2 + xy + y^2)^2 (x - y)^2 (x + y)^2;
((x^2 - xy + y^2) (x^2 + xy + y^2) (x - y) (x + y))^2;
(((x + y) (x^2 - xy + y^2)) ((x - y) (x^2 + xy + y^2)))^2;
((x^3 + y^3) (x^3 - y^3))^2;
((2^3 + 1^3) (2^3 - 1^3))^2;
((8 + 1) (8 - 1))^2;
63^2;
3969;

(a + b) (a^4 + b^4) (a - b) (a^2 + b^2);
(a - b) (a + b) (a^2 + b^2) (a^4 + b^4);
(a^2 - b^2) (a^2 + b^2) (a^4 + b^4);
(a^4 - b^4) (a^4 + b^4);
(a^8 - b^8);
(1^8 - (-2)^8)
-255

Answer 5

（x^2-xy+y^2)^2(x^2+xy+y^2)^2(x-y)^2(x+y)^2
=((x^2-xy+y^2)(x+y)(x^2+xy+y^2)(x-y))^2
=((x^3+y^3)(x^3-y^3))^2
=(x^6-y^6)^2
=(2^6-1^6)^2
=63^2
=3969

(a+b)(a^4+b^4)(a-b)(a^2+b^2)
=(a^4+b^4)(a^2+b^2)(a+b)(a-b)
=(a^4+b^4)(a^2+b^2)(a^2-b^2)
=(a^4+b^4)(a^4-b^4)
=a^8-b^8
=1^8-(-2)^8
=-255