## Diophantine equation: (x – y – z)^3 = 27 xyz

Are there any integral solutions   $(x, \; y, \; z)$   of the Diophantine equation

$(x - y - z)^3 \; = \; 27 \, x \, y \, z$

other than   $(-n, \; n, \; n)$   or such that   $x \, y \, z \; = \; 0$ ?

Solution:

Let

$x \; = \; a^3$
$y \; = \; b^3$
$z \; = \; c^3$

so that   $a^3 - b^3 - c^3 \; = \; 3 \, a \, b \, c$,   or equivalently

$(a - b - c) \,((a + b)^2 + (a + c)^2 + (b - c)^2) \; = \; 0$   ………..   (1)

since   (1)   it is equivalent to   $2 \, (a^3 - 3 \, a \, b \, c - b^3 - c^3) \; = \; 0$

Hence an infinite class of non-trivial solutions is given by

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

I don’t know whether or not there are any other solutions.

math grad - Interest: Number theory
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### 2 Responses to Diophantine equation: (x – y – z)^3 = 27 xyz

1. Paul says:

Like these?, format {x, y, z}

{-27,-8,-1}
{-27,-1,-8}
{-8,-27,1}
{-8,1,-27}
{-6,-48,6}
{-6,6,-48}
{-5,-40,5}
{-5,5,-40}
{-4,-32,4}
{-4,4,-32}
{-3,-24,3}
{-3,3,-24}
{-2,-16,2}
{-2,2,-16}
{-1,-27,8}
{-1,-8,1}
{-1,1,-8}
{-1,8,-27}
{1,-8,27}
{1,-1,8}
{1,8,-1}
{1,27,-8}
{2,-2,16}
{2,16,-2}
{3,-3,24}
{3,24,-3}
{4,-4,32}
{4,32,-4}
{5,-5,40}
{5,40,-5}
{6,-6,48}
{6,48,-6}
{8,-1,27}
{8,27,-1}
{27,1,8}
{27,8,1}