An identity form: $x_1^3+x_2^3+x_3^3+x_4^3=y_1^3+y_2^3+y_3^3+y_4^3$

Question

I found an identity form: $x_1^3+x_2^3+x_3^3+x_4^3=y_1^3+y_2^3+y_3^3+y_4^3$ as follows: $$(a+b+c)^3+a^3+b^3+c^3=(a+b)^3+(b+c)^3+(c+a)^3+(6y)^3$$

Where $abc=36y^3$

Poof of this identity is very simple. But the identity nice.

Which reference of the identity? can generalized of the identity?

$\begingroup$ see here sites.google.com/site/tpiezas/001a $\endgroup$
– Dr. Sonnhard Graubner
Jul 5, 2017 at 17:39 — Dr. Sonnhard Graubner, Jul 5, 2017 at 17:39
$\begingroup$ math.stackexchange.com/questions/1588122/… $\endgroup$
– individ
Jul 5, 2017 at 17:40 — individ, Jul 5, 2017 at 17:40
$\begingroup$ See (math.stackexchange.com/q/381111). $\endgroup$
– Jean Marie
Jul 5, 2017 at 20:43 — Jean Marie, Jul 5, 2017 at 20:43

Sam · Accepted Answer · 2017-07-28 04:25:30Z

1

Equation (shown below) has parametrization:

$(a+b+c)^3+a^3+b^3+c^3=(a+b)^3+(b+c)^3+(c+a)^3+(6y)^3$

$(y,a,b,c)=((8)(3k-4),(256),(24)(3k-4),(3)(3k-4)^2)$

For $k=2$, we get: $(3,12,64,79)^3=(15,24,67,76)^3$

answered Jul 28, 2017 at 4:25

Sam

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