Wed, 30 Nov 2022 02:44:52 +0000

One way (Easy to check), if \( x ,y. z\) are integers and
𝑥³+𝑦³=6𝑧³
Then [−𝑥(𝑦³+6𝑧³)]³+[𝑦(𝑥³+6𝑧³)]³=6[𝑧(𝑦³−x³)]³
Starting with an initial value of (17,37,21) for (x,y,z)
We can find the new three values.

This way there are infinite such solutions.
(Not all of them give all positive values but we can keep doing this till all values are positive. :)