<s>[INST] Can the cube of every perfect number be written as the sum of three cubes? I found an amazing conjecture: the cube of every perfect number can be written as the sum of three positive  cubes. The equation is
$$x^3+y^3+z^3=\sigma^3$$
where $\sigma$ is a perfect number
(well it holds good for the first three perfect numbers that is):
$${ 3 }^{ 3 }+{ 4 }^{ 3 }+{ 5 }^{ 3 }={ 6 }^{ 3 }$$
$${ 21 }^{ 3 }+{ 18 }^{ 3 }+{ 19 }^{ 3 }={ 28 }^{ 3 }$$
$${ 495 }^{ 3 }+{ 82 }^{ 3 }+{ 57 }^{ 3 }={ 496 }^{ 3 }$$
Is this what I am proposing that the cube of any perfect number can be expressed in terms of the sum of three positive cubes true?
If it is then can we prove it?
 [/INST] This is more of a comment as opposed to an answer

There is a formula for finding the values of $a, b, c, d$ in the following equation: $$a^3 + b^3 + c^3 = d^3$$ Where $$\forall x, y\in \mathbb{Z}, \ \begin{align} a &= 3x^2 + 5y(x - y), \ b = 2\big(2x(x - y) + 3y^2\big) \\ c &= 5x(x - y) - 3y^2, \ d = 2\big(3x^2 - 2y(x + y)\big) \end{align}$$
Therefore if we prove this conjecture, we prove that every perfect number $d$ must be even! I also would like to expand this theorem as well with a theorem of mine:


If $$\forall\{x, y, z\}\subset \mathbb{N}, \ 6^3 + (2x)^3 + (2y - 1)^3 = z^3$$ Then $$z \equiv \pm 1 \pmod 6$$ Where $z$ is a prime number. If $z$ is not a prime number, then $z\equiv3\pmod 6$

Examples: $$\begin{align} 6^3 + 8^3 + 1^3 &= 9^3 \qquad \ \ \ \ \mid9 &\equiv 3 \pmod 6 \\ 6^3 + 32^3 + 33^3 &= 41^3 \qquad \ \ \mid41 &\equiv 5 \pmod 6 \\ 6^3 + 180^3 + 127^3 &= 199^3 \qquad \mid199 &\equiv 1 \pmod 6 \\ 6^3 + 216^3 + 179^3 &= 251^3 \qquad \mid251 &\equiv 5 \pmod 6 \\ 6^3 + 718^3 + 479^3 &= 783^3 \qquad \mid783 &\equiv 3 \pmod 6 \\ 6^3 + 768^3 + 121^3 &= 769^3 \qquad \mid769 &\equiv 1 \pmod 6 \end{align}$$

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