若實數x,y滿足 x+y=3,x^2+y^2=7,求x^5+y^5的值

題目：

若實數x,y滿足 x+y=3,x^2+y^2=7,求x^5+y^5的值

解答：

x^2+y^2=(x+y)^2-2xy
即7=3^2-2xy
則2xy=9-7=2
xy=1
x^3+y^3=(x+y)(x^2-xy+y^2)
=3×(7-1)
=18
(x^2+y^2)(x^3+y^3)
=x^5+x^2y^3+y^2x^3+y^5
將xy=1代入上式
得x^5+y^5+y+x
=x^5+y^5+3
又(x^2+y^2)(x^3+y^3)
=7×18
=126
即
x^5+y^5+3=126
則
x^5+y^5=123