Partitions of sums of $k$ random odd/even $r$th powers from array of consecutive $r$th powers

Question

Like my previous question, I'll pose this one too with an array.

$1^r, 3^r, 5^r, 7^r, 9^r$ (all odd $r$th powers)

That's array 1. And array 2;

$2^r, 4^r, 6^r, 8^r, 10^r$ (all even $r$th powers)

Let's take the sum of random $k$ integers from each array, where $r>2$ (to eliminate Pythagorean triplets), so our $r=3, k=3$, and from Array 1, we have $(1^3 + 5^3 + 7^3)=469$ and for Array 2, $(6^3 + 10^3 + 2^3)=1224$

Now let me get to the point. I want to know, is it possible for the sum of $k$ random odd/even $r$th powers from an array of consecutive odd/even $r$th powers respectively, to be partitioned in any other way than the $r$th powers that constituted them?

I guess the answer for this must be $1$ (the initial sum of $r$th powers)...

PS I've taken separate arrays to avoid the trouble of Hardy-Ramanujan-like numbers (as in cubes) and equivalence to other powers (like $3^3 + 4^3 + 5^3=6^3$) from occuring in the sum. The effort is to keep the possible partitions as close as possible to $1$ (initial composition).

Answer 1

Here's an example with odd numbers: $$1^3+9^3+15^3+23^3=3^3+5^3+19^3+21^3$$

Answer 2

If you look at the small summary about "taxicab" numbers, you will see that there are a fair number of $(s,t)$, $(u,v)$ such that $s^3+t^3=u^3+v^3$ and $s$ and $t$ are both odd, while $u$ and $v$ are both even.

By looking at tables of Hardy-Ramanujan numbers, we can produce examples that have more cubes. For $k=4$, all we need to do is to find in the tables two numbers $A$ and $B$ which can each be expressed as the sum of two odd cubes and also of two even cubes. Then the sum $A+B$ can be expressed as the sum of four odd cubes, and also of four even cubes.