## Equation: x^3 + y^3 + z^3 – 3xyz = 0

Older blog:  x^3 + y^3 + z^3 ± (x+y+z) = 0

parametric solutions:

$x \; = \; (a-b)^3 \; + \; (a-c)^3$
$y \; = \; (b-a)^3 \; + \; (b-c)^3$
$z \; = \; (c-a)^3 \; + \; (c-b)^3$

where

$x + y + z = (a-b)^3+(a-c)^3+(b-a)^3+(b-c)^3+(c-a)^3+(c-b)^3=0$

$x^3 \; + \; y^3 \; + \; z^3$

$= \; ((a-b)^3+(a-c)^3)^3 \; + \; ((b-a)^3 \; + \; (b-c)^3)^3 \; + \; ((c-a)^3 + (c-b)^3)^3$

$= \; 3 \, ((a-b)^3 + (a-c)^3) \, ((b-a)^3 + (b-c)^3) \, ((c-a)^3 + (c-b)^3)$

$= \; 3 \,x \, y \, z$

and, let’s note the following:

$((a-b)^3+(a-c)^3)^2 \; + \; ((b-a)^3+(b-c)^3)^2 \; + \; ((c-a)^3+(c-b)^3)^2$
$= \; 6 \, (a^2+b^2+c^2-a \, b-a \, c-b \, c)^3$

$((a-b)^3+(a-c)^3)^4 \; + \; ((b-a)^3+(b-c)^3)^4 \; + \; ((c-a)^3+(c-b)^3)^4$
$= \; 18 \, (a^2+b^2+c^2-a \, b-a \, c-b \, c)^6$

$((a-b)^3+(a-c)^3)^3+((b-a)^3+(b-c)^3)^3+((c-a)^3+(c-b)^3)^3$
$= 3 (a+b-2 \, c) (2 \, a-b-c) (a-2 \, b+c) (a^2-a \, b-a \, c+b^2-b \, c+c^2)^3$

$((a-b)^3+(a-c)^3)^5+((b-a)^3+(b-c)^3)^5+((c-a)^3+(c-b)^3)^5$
$= 15 (a+b-2 \, c) (2 \, a-b-c) (a-2 \, b+c) (a^2-a \, b-a \, c+b^2-b \, c+c^2)^6$

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## About benvitalis

math grad - Interest: Number theory
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