Diophantine equation $x^3=a^2+b^2+c^2$

A positive integer is the sum of three squares iff it is not of the form $4^a (8b + 7)$. Note that $j^3 \equiv j \mod 8$ if $j$ is odd. If $j$ is odd, $(2^k j)^3 = 2^{3k} j^3$: $2^{3k}$ is a power of $4$ iff $k$ is even, and $j^3 \equiv 7 \mod 8$ iff $j \equiv 7 \mod 8$. So the only cases where $x = 2^k j$ is not a solution are when $k$ is even and $j \equiv 7 \mod 8$, i.e. when $x$ itself is not a sum of three squares.

For any positive integer $k>2$, the kth power of the sum of two squares is the sum of three squares,

$$x^2+y^2+z^2 = (a^2+b^2)^k$$

thus,

$$a^2(a^2+b^2)^2 + (2ab^2)^2 + (a^2-b^2)^2b^2 = (a^2+b^2)^3$$

$$(a^4-b^4)^2 + (4a^2b^2)^2 + (2ab(a^2-b^2))^2 = (a^2+b^2)^4$$

$$a^2(a^2+b^2)^4 + (4a^3b^2-4ab^4)^2 + (a^4b-6a^2b^3+b^5)^2 = (a^2+b^2)^5$$

and so on. Similarly, for $k\geq2$, the kth power of the sum of three squares is itself the sum of three squares. Thus,

$$(a^2-b^2-c^2)^2 + (2ab)^2 + (2ac)^2 = (a^2+b^2+c^2)^2$$

$$(a^3-3ab^2-3ac^2)^2 + b^2(-3a^2+b^2+c^2)^2 + c^2(-3a^2+b^2+c^2)^2 = (a^2+b^2+c^2)^3$$

and so on.

Note: For $k>2$, this does not give all solutions, but at least you have formulas. For more details, see Sums of Three Squares.

The simple formula for the equation:

$$x^2+y^2+z^2=q^3$$

You can write this:

$$x=3(p-k-t)(p^2+2k^2-2kt+2t^2)s^3$$

$$y=3(p-k+2t)(p^2+2k^2-2kt+2t^2)s^3$$

$$z=3(p+2k-t)(p^2+2k^2-2kt+2t^2)s^3$$

$$q=3(p^2+2k^2-2kt+2t^2)s^2$$

$p,k,t$ - integers asked us.

If there is a one simple solution, it will be necessary to reduce to the corresponding $s$.

There are other formulas. But they are bulky. Don't know whether it makes sense to write them.

https://artofproblemsolving.com/community/c3046h1054692_the_squares_with_cube_2

Sum of three squares in two different ways is equal to a cube :

$$(x^2+y^2)^3=(x^3+xy^2)^2+(2xy^2)^2+(x^2y-y^3)^2 =(x^3-xy^2)^2+(2x^2y)^2+(x^2y+y^3)^2$$

$$[3(x^2+y^2)^2]^3=(x^6+6x^5y-15x^4y^2-20x^3y^3+15x^2y^4+6xy^5-y^6)^2+(x^6-16x^5y+3x^4y^2+3x^2y^4+16xy^5+y^6)^2+(5x^6+2x^5y-11x^4y^2+36x^3y^3+11x^2y^4+2xy^5-5y^6)^2= (x^6 + 14x^5y - 31x^4y^2 -36x^3y^3 +31x^2y^4 +14xy^5 -y^6)^2+(x^6 +16x^5y + 3x^4y^2 +3x^2y^4 - 16xy^5 +y^6)^2+(5x^6 - 6x^5y - 27x^4y^2 +52x^3y^3 + 27x^2y^4 - 6xy^5 -5y^6)^2$$

Sum of four squares in three different ways is equal to a cube :

$$(x^2+y^2+1)^3=(2x^2+2y^2)^2+(x^3+xy^2-x)^2+(x^2y-y+y^3)^2+(x^2+y^2+1)^2=(2x^2y+2y)^2+(x^3-xy^2+x)^2+(x^2y+y+y^3)^2+(x^2-y^2+1)^2=(2x+2xy^2)^2+(x^3+xy^2+x)^2+(x^2y-y-y^3)^2+(x^2-y^2-1)^2$$

Sum of five squares is equal to cube can be written as more than three different ways.