a4(b2-c2)+b4(c2-a2)+c4(a2-b2)的解

原式=a4b2-a4c2+b4c2-b4a2+c4a2-c4b2=a2b2(a2-b2)-c2(b4-a4)+c4(a2-b2)=a2b2(a+b)(a-b)-c2(b2+c2)(b+a)(b-a)+c4(a+b)(a-b)=(a+b)(a-b)(a2b2-b2c2-a2c2+c4)=(a+b)(a-b)(b2-c2)(a2-c2)=(a+b)(a-b)(b+c)(b-c)(a+c)(a-c)...