A 3 B 3 C 3 3abc A B C A 2 B 2 C 2 Ab Bc Ca Prove
A 3 b 3 c 3 3abc a b c a 2 b 2 c 2 ab bc ca and we get.
A 3 b 3 c 3 3abc a b c a 2 b 2 c 2 ab bc ca prove. A 3 b 3 c 3 ab a b ac a c bc b c since most problems involving am gm are symmetric. A 3 b 3 c 3 3abc a b c a 2 b 2 c 2 ab bc ca where a 2a b 3b and c 5c now apply values of a b and c on the l h s of identity i e. By assumption a 3 b 3 c 3 3abc so the left hand side is 0. Now suppose a 2 b 2 c 2 ab ac bc 0.
V 1 a 2 bc c v 2 b 2 ac a v 3 c 2 ab b. A 3 b 3 c 3 3abc a b c a 2 b 2 c 2 ab ac bc on the right side you have to multiply everything out since there s nothing you can do on the left hand side. A 3 b 3 c 3 3abc a b c a 2 b 2 c 2 ab ac bc. Therefore a b c a 2 b 2 c 2 ab ac bc 0.
Factorise a 3 b 3 c 3 3abc get the answer to this question by visiting byju s q a forum. Recommend 0 comment 0 ask a question. Please allow me to complete the reasoning proposed with above post let us further consider the triangle of vertices. You can re write this as.
2a 3b 5c 2a 2 3b 2 5c 2 2a 3b 3b 5c 5c 2a expand the. I worked this one out but i ll wait for a confirmation from you that this is what you want proved before posting my work. What must be subtracted from 4x 4 2x 3 6x 2 2x 6 so.