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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

$a^4+b^4+c^4=(a+b+c)(a^3+b^3+c^3-a^{2}b-a^{2}c-a{b}^2-a{c}^2-b^{2}c-b{c}^2)+2(a^2{b}^2+a^2{c}^2+b^2{c}^2)$

$a^5+b^5+c^5=(a+b+c)(\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r})-5(a^2{b}^{2}c+a^{2}b{c}^2+a{b}^2{c}^2)$

$a^6+b^6+c^6=(a+b+c)(\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r})-2(a^3{b}^3+a^3{c}^3+b^3{c}^3)+9a^2{b}^2{c}^2$

$a^7+b^7+c^7=(a+b+c)(\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r})+7(a^3{b}^{3}c+a^{3}b{c}^3+a{b}^3{c}^3)$

$a^8+b^8+c^8=(a+b+c)(\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r})+2(a^4{b}^4+a^4{c}^4+b^4{c}^4)-16(a^3{b}^3{c}^2+a^3{b}^2{c}^3+a^2{b}^3{c}^3)$

$a^9+b^9+c^9=(a+b+c)(\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r})-9(a^4{b}^{4}c+a^{4}b{c}^4+a{b}^4{c}^4)+30a^3{b}^3{c}^3$

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)