4(2x^2-x+1)(x^2-2x+3)-(3x^2-3x+4)^2
=4(2x^2-x+1)(x^2-2x+3)-[(2x^2-x+1)+(x^2-2x+3)]^2
=4(2x^2-x+1)(x^2-2x+3)-(2x^2-x+1)^2-(x^2-2x+3)^2-2(2x^2-x+1)(x^2-2x+3)
=-(2x^2-x+1)^2-(x^2-2x+3)^2+2(2x^2-x+1)(x^2-2x+3)
=-[(2x^2-x+1)^2+(x^2-2x+3)^2-2(2x^2-x+1)(x^2-2x+3)
=-[(2x^2-x+1)-(x^2-2x+3)]^2
=-(x^2+x-2)^2
=-(x+2)^2(x-1)^2
参考：
令a=2x²-x+1
b=x²-2x+3
则3x²-3x+4=a+b
所以原式=4ab-(a+b)²
=4ab-a²-2ab-b²
=-(a²-2ab+b²)
=-(a-b)²
=-(2x²-x+1-x²+2x-3)²
=-(x²+x-2)²
=-(x-1)²(x+2)²
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