algorithm for positive integer solutions of equation $a^3+b^3=22c^3$ This is a look-a-like to Fermat's last theorem for $n=3$, but it has solutions!  I believe that its solution requires knowledge of the techniques of algebraic or analytic number theory which I don't have. 

Might have found something to generate more from an initial solution.
Suppose $a^3+b^3=22c^3$. If want the smallest such $a$ and $b$, they have to be relatively prime to each other and to $c$. $a^3+b^3=(a+b)(a^2+b^2-ab)$. Via trial and error, $a^2+b^2-ab$ cannot be divisible by 11 unless both $a$ and $b$ are, which is not allowed. Combined, this tells us $a+b$ is divisible by $22$.
After substitution: $a^3+(22k-a)^3=22c^3$. After expanding, cancelling the term cubic in $a$ and finding the discriminant of the resulting quadratic, one finds $c^3=11^2k^3+3n^2k=k(11^2k^2+3n^2)$ where $a=11k+n$ and $b=11k-n$. So $k$ and $n$ must be relatively prime in order for $a$ and $b$ to be relatively prime. This leads us to consider two scenarios.
If $k$ is divisible by 3, it can be shown that it must also be divisible by 9. Then $c$ must also be divisible by 3. If $c=3p$ and $k=9q$, then $p^3=11^2\cdot27\cdot q^3+qn^2=q[3\cdot(33)^2q^2+n^2]$.  Since $q$ and $n$ are relatively prime then q and the term in square brackets must also be relatively prime. If the product of two relatively prime numbers is a perfect cube, then the factors themselves must be perfect cubes.
This suggests there is a $x$ so that  $x^3=3\cdot(33)^2q^2+n^2$.
The form of the equation remains largely unchanged upon multiplication by 27.
$27x^3=81\cdot(33)^2q^2+27n^2\implies  (3x)^3=(33)^2(9q)^2+3(3\cdot n)^2$ Again we have an expression of the form a perfect cube is the sum of three times a perfect square plus some other perfect square. A similar  situation occurs if $k$ is not taken to be divisible by 3.
By arguments above we also have that a perfect cube has that form if we don't take that 3 divides $k$.
So possible solutions have the form $x^3 = 3\cdot(33)^2r^6+n^2$ where $q=r^3$ or $x^3=11^2\cdot r^6+3n^2$ where $k=r^3$.
We have an example of the first where $r=6$ and $n=4085$. This leads to $a=17299, b=25579, c=9954$, a solution mentioned elsewhere.
