解
x²-3x=2
原式=[(x-1)³-(x²-1)]/(x-1)
=[(x-1)³-(x-1)(x+1)]/(x-1)
=(x-1)²-(x+1)
=x²-2x+1-x-1
=x²-3x
=2
（(x-1)³-x²+1）/(x-1)=(x-1)(x²-2x+1-x-1)/(x-1)=x²-3x
因为x²-3x-2=0,即x²-3x=2
带入得
（(x-1)³-x²+1）/(x-1)=(x-1)(x²-2x+1-x-1)/(x-1)=x²-3x=2
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