Wikipedia:Reference desk/Archives/Mathematics/2019 October 16

= October 16 =

Combinatorial aspects in identities involving sum of n cubes divisibility
Starting from the previous cases of identities involving sum of cubes of positive integers (3 and 4 cubes) divided by the sum of numbers present in cubes, what is the general formula for the sum of n cubes divided by the sum of n numbers? What is the general form of sum of products of succesive cases of the form 3(abc + abd + bcd + acd) (4 choose 3 or 4C3 or C(4,3)) encountered in the case of four cubes compared with sum of 5 cubes, 6 cubes, and so on. (C(n,3)?)..?--109.166.137.3 (talk) 18:36, 16 October 2019 (UTC)


 * One form of the general relationship that captures any number of variables is $$p_3 = e_1 \cdot (p_2 - e_2) + 3e_3$$, where p is the power sum symmetric polynomial and e is the elementary symmetric polynomial. --JBL (talk) 02:00, 17 October 2019 (UTC)


 * Before anyone asks, higher powers are more complicated, but some analogous identities do exist. For example
 * $$a^4 + b^4 + c^4 = (a+b+c)(a^3+b^3+c^3-a^2b-a^2c-ab^2-ac^2-b^2c-bc^2) + 2(a^2b^2+a^2c^2+b^2c^2)$$
 * $$a^5 + b^5 + c^5 = (a+b+c)({\rm some\,factor}) - 5(a^2b^2c+a^2bc^2+ab^2c^2)$$
 * $$a^6 + b^6 + c^6 = (a+b+c)({\rm some\,factor}) - 2(a^3b^3+a^3c^3+b^3c^3)+9a^2b^2c^2$$
 * $$a^7 + b^7 + c^7 = (a+b+c)({\rm some\,factor}) + 7(a^3b^3c+a^3bc^3+ab^3c^3)$$
 * $$a^8 + b^8 + c^8 = (a+b+c)({\rm some\,factor}) + 2(a^4b^4+a^4c^4+b^4c^4)-16(a^3b^3c^2+a^3b^2c^3+a^2b^3c^3)$$
 * $$a^9 + b^9 + c^9 = (a+b+c)({\rm some\,factor}) - 9(a^4b^4c+a^4bc^4+ab^4c^4)+30a^3b^3c^3$$
 * with the formulas getting longer and longer. With more variables yet more complicated identities exist; I'm not going to work them out though. --RDBury (talk) 09:14, 17 October 2019 (UTC)