Numerical puzzle I'm stuck here with some numerical rebus -
Given: $A^2=BC, A^3=CA$
Find: $A+B+C$


*

*$13$

*$12$

*$11$

*$10$


(only one correct solution)
Note that letters represent digits.
I can't think of any idea to solve this one, and according to the book from which this question was taken, it is possible to solve in one minute. I will appreciate any idea.

Assuming $A^2=10B+C$ and $A^3=10C+A$
We get $A^3<100\implies A<5$
As the last digits of $A$ and $A^3$ are same, $A$ must be $0,1$ or $4$
$A=0\implies C=B=0$
$A=1,$ 
from the 1st equation $1^2=10B+C\implies B=0,C=1$
from the 2nd equation $1^3=10C+1\implies C=0$ so $A\ne 1$ as it would make the system inconsistent.
$A=4\implies C=6,B=1 $
