已知x-y=1,x三次方-y三次方=2，求 x四次+y四次 和 x五次-y五次

解：∵x^3-y^3=（x-y）（x^2+xy+y^2）=2，
x-y=1，
x^3-y^3=（x-y）（x^2+xy+y^2）=2，
又∵x^2-2xy+y^2=1，与上式联立得：
xy=1/3，x^2+y^2=5/3，
故x^4+y^4=（x^2+y^2）2-2x^2y^2=23/9，

又x^5-y^5=x^5-x^4y+x^4y-xy^4+xy^4-y^5=x^4（x-y）+xy（x^3-y^3）+y^4（x-y），
将x-y=1，xy=1/3，x^3-y^3=2代入，
可得x^5-y^5=29/9，
故答案为23/9、29/9．

x-y=1, 两边平方有 x^2 -2xy+y^2=1
x^3-y^3=(x-y)(x^2+xy+y^2)=x^2+xy+y^2=2
以上两式联立有xy=1/3, x^2+y^2=5/3
x^4+y^4 = (x-y)(x^3-y^4) +xy(x^2+y^2)= 1*2+ 1/3*(5/3)= 2+ 5/9 = 23/9
x^5-y^5 = (x^4+y^4)(x-y) -xy^4 +x^4y=23/9 +xy(x^3-y^3)= 23/9 + 2*1/3=29/9

x-y=1,
∵x³-y³=2
∴(x-y)(x^2+xy+y^2)=2
∵x-y=1
∴x^2+xy+y^2=2 (1)
x-y=1==> x^2-2xy+y^2=1 (2)
(1)-(2) ==> 3xy=2,xy=1/3
x^2+y^2 =5/3
∴x^4+y^4
=(x^2+y^2)^2-2x^2y^2
=(5/3)^2-2(1/3)^2=25/9-2/9=23/9
x^5-y^5=x^5-x^3y^2+x^3y^2-y^5
=x^3(x^2-y^2)+y^2(x^3-y^3)
=x^3(x+y)(x-y)+y^2(x-y)(x^2+xy+y^2)
=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)
=(x-y)[(x^4+y^4)+x^2y^2+xy(x^2+y^2)]
=1(23/9+1/9+1/3*5/3)=29/9