Question

已知x+y=1,x^2+y^2=3,求(1)x^3+y^3的值，（2）x^5+y^5的值

Answer 1

由已知得，xy=[(x+y)^2-(x^2+y^2)]/2=-1
(1)x^3+y^3=(x^2+y^2)(x+y)-xy^2-x^2y=(x^2+y^2)(x+y)-xy(X+y)=3*1+1=4
(2)x^4+y^4=(x^3+y^3)(x+y)-xy(x^2+y^2)=7
而x^2y^2=[（x^2+y^2)^2-(x^4+y^4)]/2=1
x^5+y^5=(x^3+y^3)*(x^2+y^2)-x^2y^2(X+y)=11

Answer 2

（1）
x^3+y^3=(x+y)(x^2-xy+y^2)
=x^2-xy+y^2
=3-xy
x^2+y^2=3
(x+y)^2-2xy=3
1-2xy=3
xy=-1
所以，原式=3-（-1）=4
（2）原式=(x^2+y^2)(x^3+y^3)-x^2*y^2(x+y)
=3*4-(-1)^2
=11

Answer 3

用已知式子求出XY=-1后把两个已知的式子相乘，化简第一问是4。第二问把已知第一个式子乘两次已知第二式子化简得11