原式=x(x+1)[(x-2)(x+3)]+8
=(x²+x)[(x²+x)-6]+8
=(x²+x)²-6(x²+x)+8
=(x²+x-2)(x²+x-4)
=(x+2)(x-1)(x²+x-4)
x(x-2)(x+3)(x+1)+8
=x(x+1)(x-2)(x+3)+8
=(x^2+x)(x^2+x-6)+8
=(x^2+x)^2-6(x^2+x)+8
=(x^2+x-4)(x^2+x-2)
=(x^2+x-4)(x+2)(x-1)
答案应为：
原式=x(x+1)(x-2)(x+3)+8
=(x²+x)(x²+x-6)+8
=(x²+x)^2-6(x²+x)+8
=(x²+x-4)(x²+x-2)
=(x²+x-4)(x+2)(x-1)
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