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?

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

