The equation $a^3 + b^3 = c^2$ has solution $(a, b, c) = (2,2,4)$. The equation $a^3 + b^3 = c^2$ has solution $(a, b, c)  = (2,2,4).$ Find 3 more solutions in positige integers.  [hint.  look for solutions of the form $(a,b,c) = (xz,  yz,  z^2)$
Attempt:
So I tried to use the hint in relation to the triple that they gave that worked.   So I observed that the triple that worked $(2,2,4)$ could also be written as $$2(1,1,2)$$  So i attempted to generalize it to $$n(1,1,2)$$ I thought this may work because I am using the "origin" of where the triple $(2,2,4)$ came from.  This did not work unfortunately.  So I am asking what else could I comsider to devise some system to find these triples? Because I highly doubt they're asking me to just shoot in the dark and pick random numbers

Here is an
infinite number of solutions:
For any positive integer $c$,
$x = (1+c^{3})$,
$y 
=c(1+c^{3})$,
and
$z = (1+c^{3})^2
$.
Check:
$x^3+y^3
=(1+c^3)^3+(c(1+c^3))^3
=(1+c^3)^3(1+c^3)
=(1+c^3)^4
$
and
$z^2
=((1+c^{3})^2)^2
=(1+c^{3})^4
$.
This is gotten from the following,
which is a re-creation
of some algebra of mine
from many years ago:
To find solutions to
$x^n+y^n = z^m$
where
$(n, m) = 1$.
Since $(n, m) = 1$,
there are $a$ and $b$
such that
$am-bn = 1$
or
$am=bn + 1$.
If
$x = u^b$,
$y = v^b$,
$z = w^a$,
then
$u^{bn}+v^{bn}
=w^{am}
=w^{bn+1}
$.
Let
$v = cu$.
Then
$u^{bn}+v^{bn}
=u^{bn}+(cu)^{bn}
=u^{bn}+c^{bn}u^{bn}
=u^{bn}(1+c^{bn})
$.
If
$w = 1+c^{bn}$,
we want
$w^{bn+1}
=w\,w^{bn}
=wu^{bn}
$
or
$u = w$.
Therefore,
a solution is
$x = (1+c^{bn})^b$,
$y = cu
=c(1+c^{bn})^b$,
and
$z = (1+c^{bn})^a
$.
If
$n=3$ and $m=2$,
we want
$2a-3b = 1$.
This is satisfied by
$a=2, b=1$.
We get
$x = (1+c^{3})$,
$y 
=c(1+c^{3})$,
and
$z = (1+c^{3})^2
$.
