いろいろ (a-b)^2 (b-c)^2 (c-a)^2 formula 187091-(a+b+c)(a^2+b^2+c^2-ab-bc-ca) formula
Please allow me to complete the reasoning proposed with above post ― Let us further consider the triangle of vertices v 1 ≡ (a 2 bc, c), v 2 ≡ (b 2 ac, a), v 3 ≡ (c 2 ab, bAug 07, 04 · 18 maverick said Use the identity for a 2 b 2 c 2 − a b − b c − c a a 2 b 2 c 2 − a b − b c − c a = 1 2 ( a − b) 2 ( b − c) 2 ( c − a) 2 The right hand side is always greater than or equal to zero (equality in the case a = b = c) This proves the resultApr 25, 16 · In general, there are only a couple expressions like a^22abb^2 and a^2b^2 that factor in such a simple manner You can even do the same thing for a^2b^2=c^2 and there are a number of solutions like the 3,4, 5 and 5,12, 13 right triangles Law Of Cosines Wikipedia (a+b+c)(a^2+b^2+c^2-ab-bc-ca) formula