以知x+y=-3 x的三次方+y的三次方=-18 求x的七次方+y的七次方等于多少?

-843

x^3+y^3
=(x+y)(x^2-xy+y^2)
=(x+y)[(x+y)^2-3xy]
=(x+y)^3-3xy(x+y)
=-27+9xy=-18
xy=1
x^7+y^7
=(x^3+y^3)(x^4+y^4)-x^3y^4-x^4y^3
=-18(x^4+y^4)-x^3y^3(x+y)
=-18(x^4+y^4)+3
x^4+y^4=(x^3+y^3)(x+y)-x^3y-xy^3
=54-(x^2+y^2)
=54-[(x+y)^2-2xy]
=54-(9-2)
=47
x^7+y^7=-18*47+3=-843
ok咯

x^3+y^3=(x+y)(x^2-xy+y^2)
(x^2-xy+y^2)=(x^3+y^3)/(x+y)=-18/(-3)=6
(x+y)^2=x^2+2xy+y^2=(x^2-xy+y^2)+3xy
(-3)^2=6+3xy
xy=1
(x^3+y^3)*(x+y)^2=x^5+y^5+2*x^4*y+2*x*y^4+x^3*y^2+x^2*y^3
=(x^5+y^5)+2xy(x^3+y^3)+x^2*y^2*(x+y)
=(x^5+y^5)-36-3
x^5+y^5=(-18)*(-3)^2+36+3=-123
(x^3+y^3)^2*(x+y)=x^7+2*x^4*y^3+x*y^6+y^7+2*x^3*y^4+x^6*y
=(x^7+y^7)+(2*x^4*y^3+2*x^3*y^4)+(x*y^6+x^6*y
)
=(x^7+y^7)+2*(x^3*y^3)(x+y)+xy(x^5+y^5)
=(x^7+y^7)-6-123
x^7+y^7=(-18)^2*(-3)+6+123=-843

x^3+y^3
=(x+y)(x^2-xy+y^2)
=(x+y)[(x+y)^2-3xy]
=(x+y)^3-3xy(x+y)
=-27+9xy=-18
xy=1
x^7+y^7
=(x^3+y^3)(x^4+y^4)-x^3y^4-x^4y^3
=-18(x^4+y^4)-x^3y^3(x+y)
=-18(x^4+y^4)+3
x^4+y^4=(x^3+y^3)(x+y)-x^3y-xy^3
=54-(x^2+y^2)
=54-[(x+y)^2-2xy]
=54-(9-2)
=47
x^7+y^7=-18*47+3=-843