Question

若x^2-y-2=0,y^2-x-2=0,(x≠y),求x^3-2xy+y^3的值

Answer 1

若
x²-y-2=0①
y²-x-2=0②
x≠y③
①-②→
(x-y)(x+y-1)=0
x+y=1
求
x³-2xy+y³
=x(x²-y)+y(y²-x)
=2(x+y)
=2

Answer 2

解:方程组为x²-y-2=0，y²-x-2=0；化为

x²-y=2，y²-x=2；则x³-2xy＋y³=x³-xy＋y³-xy=x(x²-y)＋y(y²-x)=2x+2y=2(x+y)

∵x²-y=2，y²-x=2 ∴有x²-y=y²-x，x²-y²＋x-y=0，(x-y)(x＋y＋1)=0 又∵x≠y ∴x＋y＋1=0，x＋y=-1 ∴x³-2xy＋y³=-2

Answer 3

解：X²-y-2=0①,y²-X-2=0②,(X≠y),①-②得：X²-y²+X-y=0,(X-y)(X+y+1)=0，因为X≠y,即X-y≠0,则X+y+1=0,X+y=-1
X³-2Xy+y³
=X(X²-y)+y(y²-X)
=2X+2y
=2(X+y)
=-2

Answer 4

解:方程为x²-y-2=0，y²-x-2=0；化为x²-y²+x-y=0，

(x-y)(x+y)+x-y=0，(x-y)(x+y+1)=0 ∵x≠y ∴x+y=-1

∵x³-2xy+y³=x³-xy+y³-xy=x(x²-y)+y(y²-x)=2x+2y=

-2