1.求(3+1)(3^2+1)(3^4+1)(3^8+1)的值.

题目

1.求(3+1)(3^2+1)(3^4+1)(3^8+1)的值.
2.已知a^3+a^2+a+1=0,那么求a^2008+2a^2000+5a^1996的值.
3.已知x+y=10,x^3+y^3=100,求x^2+y^2的值.

答案

(3+1)(3^2+1)(3^4+1)(3^8+1)
=1/2 *(3-1)*(3+1)(3^2+1)(3^4+1)(3^8+1)
=1/2*(3^2-1)*(3^2+1)(3^4+1)(3^8+1)
=1/2*(3^4-1)(3^4+1)(3^8+1)
=1/2*(3^8-1)(3^8+1)
=1/2*(3^16-1)
=21523360
a^3+a^2+a+1=0
a^2(a+1)+(a+1)=0
(a^2+1)(a+1）=0
所以a=-1
所以a^2008+2a^2000+5a^1996
=1+2+5=8
x^3+y^3=(x+y)(x^2+y^2-xy)
10*(x^2+y^2-xy)=100
x^2+y^2-xy=10
x^2+y^2-[(x+y)^2-(x^2+y^2)]/2=10
(x^2+y^2)-[100-(x^2+y^2)]/2=10
所以x^2+y^2=40