x³+x+2005=0 ①
y³+3y²+4y+2007=0
(y³+y²)+2(y²+2y+1)+2005=0
y²(y+1)+2(y+1)²+2005=0
(y+1)(y²+2y+2)+2005=0
(y+1)³+(y+1)+2005=0 ②
根据①、②式,得x=y+1
故x-y=1
x³+x+2005=0 ①
y³+3y²+4y+2007=0
(y³+y²)+2(y²+2y+1)+2005=0
y²(y+1)+2(y+1)²+2005=0
(y+1)(y²+2y+2)+2005=0
(y+1)³+(y+1)+2005=0 ②
根据①、②式,得x=y+1
故x-y=1