求证:

(1/a-b)+(1/b-c)+(1/c-a)]*[(1/a-b)+(1/b-c)+(1/c-a)
= [1/(a-b)(a-b)]+[1/(a-b)(b-c)]+[1/(a-b)(c-a)]
+[1/(b-c)(b-c)]+[1/(b-c)(a-b)]+[1/(b-c)(c-a)]
+[1/(c-a)(c-a)]+[1/(c-a)(a-b)]+[1/(c-a)(b-c)]
= 1/(a-b)(a-b)+1/(b-c)(b-c)+1/(c-a)(c-a)
+[2/(a-b)(b-c)]+[2/(a-b)(c-a)]+[2/(b-c)(c-a)]
=1/(a-b)(a-b)+1/(b-c)(b-c)+1/(c-a)(c-a)
+{[2(b-c)+2(c-a)+2(a-b)]/[(a-b)(b-c)(c-a)]}
=1/(a-b)(a-b)+1/(b-c)(b-c)+1/(c-a)(c-a)
+(2b-2c+2c-2a+2a-2b)/[(a-b)(b-c)(c-a)]
=1/(a-b)(a-b)+1/(b-c)(b-c)+1/(c-a)(c-a)
+(0)/[(a-b)(b-c)(c-a)]
=1/(a-b)(a-b)+1/(b-c)(b-c)+1/(c-a)(c-a)