Répondre :
1/ (a-b)(a-c)+1/(b-c)(b-a)+1/(c-a)(c-b)
=1/ (a-b)(a-c)-1/(b-c)(a-b)+1/(a-c)(b-c)
=(b-c)/ ((a-b)(a-c)(b-c))-(a-c)/ ((a-b)(a-c)(b-c)+(a-b)/((a-b)(a-c)(b-c)
=(b-c-a+c+a-b) ((a-b)(a-c)(b-c)
=0
=1/ (a-b)(a-c)-1/(b-c)(a-b)+1/(a-c)(b-c)
=(b-c)/ ((a-b)(a-c)(b-c))-(a-c)/ ((a-b)(a-c)(b-c)+(a-b)/((a-b)(a-c)(b-c)
=(b-c-a+c+a-b) ((a-b)(a-c)(b-c)
=0