(x+1)³-(x-1)³ =[(x+1)-(x-1)][(x+1)²+(x+1)(x-1)+(x-1)²] =2(3x²+1) 由已知 x²+x-1=0得 x²=1-x代入上式得 原式=2(3x²+1)=2(3-3x+1)=8-6x...