设A=2分之1m=1,B=2分之1m+2,C=2分之1m+3,求代数式a平方+2ab+b平方-2ac-2bc+c平方的值

a^2+2ab+b^2-2ac-2bc+c^2
=(a+b)^2-2c(a+b)+c^2
=(a+b-c)^2
=[(m+1)/2+(m+2)/2-(m+3)/2]^2
=(m/2)^2
=m^2/4
-3x^3-12x^2+36x
=-3x(x^3+4x-12)
=-3x(x+6)(x-2)
(a^2-b^2)^2+3（a^2-b^2）-18
=(a^2-b^2+6)(a^2-b^2-3)
(m^2+3m)^2-8(m^2+3m)-20
=(m^2+3m-10)(m^2+3m+2)
=(m+5)(m-2)(m+1)(m+2)
(y^2+3y)-(2y+6)^2
=-4(y+3)^2+y(y+3)
=-(4y+12-y)(y+3)
=-(3y+12)(y+3)
=-3(y+4)(y+3)
-a+2a^2-a^3
=-a(a^2-2a+1)
=-a(a-1)^2
4a^2-9b^2+c^2-4ac
=4a^2-4ac+c^2-9b^2
=(2a-c)^2 -(3b)^2
=(2a-c+3b)(2a-c-3b)