设（t+1）3=x，y=t2+t+2，则原式=[2（t2+t+2）-3（1+3t+3t2+t3）][（t2+t+2）-（1+3t+3t2+t3）]-[（t+1）3]2=（2y-3x）（y-x）-x2=2x2-5xy+2y2=（2x-y）（x-2y）=[2（t3+3t2+3t+1）-（t2+t+2）][（t3+3t2+3t+1）-2（...