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 解：原式$=[(a + b + c)+(a - b - c)][(a + b + c)-(a - b - c)]=(a + b + c+a - b - c)(a + b + c - a + b + c)=2a(2b + 2c)=4ab + 4ac$

 解：原式$=[(x - 3y + 2z)+(x + 3y - 2z)][(x - 3y + 2z)-(x + 3y - 2z)]=(x - 3y + 2z+x + 3y - 2z)(x - 3y + 2z - x - 3y + 2z)=2x(-6y + 4z)=-12xy + 8xz$

 解：原式$=2023^{2}-(2023 - 2)\times(2023 + 2)=2023^{2}-(2023^{2}-2^{2})=2023^{2}-2023^{2}+2^{2}=4$

 解：原式$=[(2m - 3)(2m + 3)]^{2}=(4m^{2}-9)^{2}=16m^{4}-72m^{2}+81$

 解：原式$=[(3a - 1)+2b][(3a - 1)-2b]=(3a - 1)^{2}-(2b)^{2}=9a^{2}-6a + 1-4b^{2}$

 解：因为$(a + 3)(b + 3)=20，$所以$ab + 3a + 3b + 9 = 20，$即$ab + 3(a + b)=11。$因为$a + b = 3，$所以$ab+3\times3 = 11，$所以$ab = 2。$所以原式$=a^{2}+2ab + b^{2}+3ab=(a + b)^{2}+3ab=3^{2}+3\times2=9 + 6 = 15$

 解：因为$(a - b)^{2}=(a + b)^{2}-4ab=3^{2}-4\times2=9 - 8 = 1，$所以$a - b=\pm1$

 解：根据题意，得$(a + b)^{4}=a^{4}+4a^{3}b + 6a^{2}b^{2}+4ab^{3}+b^{4}，$所以$x^{4}-12x^{3}+54x^{2}-108x + 81=x^{4}+4x^{3}\cdot(-3)+6x^{2}\cdot(-3)^{2}+4x\cdot(-3)^{3}+(-3)^{4}=(x - 3)^{4}，$所以$(x - 3)^{4}=1，$所以$x - 3 = 1$或$x - 3=-1，$所以$x = 2$或$4$