①-1/64x^3+64y^3,利用a^3-b^3=(a-b)(a^2+ab+b^2),
-1/64x^3+64y^3=[-1/(4x)]^3+[1/(4y)]^3=[1/(4y)]^3-[1/(4x)]^3=[1/(4y)-1/(4x0][1/(4y)^2+1/(4x)^2+1/(16xy)]
②a^7-a=a(a^6-1)=a(a^2-1)(a^+a+1)=(a-1)a(a+1)(a^+a+1),其中(a-1)a(a+1)为三个连续整数相乘,因此能被6整除
③是这个意思么、?ax^2+ay^2+2axy-ab^2
=a(x^2+2xy+y^2-b^2)=a[(x+y)^2-b^2]=a(x+y+b)(x+y-b)
④已知a=1/20x+20,b=1/20x+19,c=1/20+21,代数式a+b+c-ab-bc-ac求不出来,但是
a^2+b^2+c^2-ab-bc-ac能够求出来
a^2+b^2+c^2-ab-bc-ac=1/2×（2a^2+2b^2+2c^2-2ab-2ac-2bc）=1/2×[（a^2-2ab+b^2）+（a^2-2ac+c^2）+（b^2-2bc+c^2）]=1/2×[(a-b)^2+(a-c)^2+(b-c)^2]
一题目条件有,a-b=1,a-c=-1,b-c=-2
故a^2+b^2+c^2-ab-bc-ac=1/2×[(a-b)^2+(a-c)^2+(b-c)^2]=1/2×（1+4+1）=3