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?

It can be shown that if $\sigma$ is an even perfect number then $\sigma^3$ can be expressed as a sum of three positive cubes.  Since there are no known odd perfect numbers, this also means that the cubes of all known perfect numbers can be so expressed. But it doesn't guarantee that every perfect number can be so expressed, since it hasn't been proved that there are no odd perfect numbers (although that seems very likely - see here).
To prove the proposition for the even case, first note that (as pointed out in Thomas Andrews' comment) an even perfect number $\sigma$ must be of the form $2^{p-1}(2^p-1)$ where $p$ and $2^p-1$ are prime (see Mathworld, equation (15) and following paragraph). Hence any $\sigma$ with $p>8$ has $2^8$ as a factor. But:
$$(2^8)^3=256^3=9^3+183^3+220^3$$
Hence for any such $\sigma$:
$$\sigma^3= (2^82^{p-9}(2^p-1))^3=(9(2^{p-9})(2^p-1))^3+(183(2^{p-9})(2^p-1))^3+(220(2^{p-9})(2^p-1))^3$$
It remains to show that the proposition holds when $p\leq8$. But the only such cases are $p=2,3,5,7$ yielding respectively $\sigma=6,28,496,8128$ which have already been  dealt with in the question and, for $8128$, in the answer by Tito Piezas III.
