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. 
 A: By a theorem of Satgé, there are rational solutions to,
$$x^3+y^3 = 2p$$
if $p \equiv 2\; \text{(mod 9)}$, of which your $p=11$ qualifies. If there is an initial non-zero integer solution to,
$$x^3+y^3 = cz^3$$
then infinitely many subsequent ones can be found as,
$$(-x^4 - 2x y^3)^3 + (2x^3 y + y^4)^3 = c (-x^3z + y^3z)^3\tag1$$
Positive $x,y,z$ will appear every few iterations. However, this identity will not give all solutions. See this related post.
P.S. By the way, what's the initial solution to $x^3+y^3=22z^3$?
A: I believe that the Marty Ellis solution is in fact an initial solution. Not being an expert in Diophantine equations it just appeared large to me, but Louis de Branges, who found solutions years ago, confirmed to me that his original generation solution was indeed of this size.  This along with Tito Piezas' formula for generation of infinitely more solutions satisfies me that my original questions are answered. Mike.  
A: 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.
