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.
 A: 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 $
A: From $A^3 = CA$, we know $A = 3$ or $A = 4$ (since $A^3$ must be two digits). Testing both, we see only $A = 4$ works, so $A^3 = 64 \rightarrow C = 6$
From $A^2 = BC$ and $A = 4$, $C = 6$, we see that $B = 1$. Thus, $A+B+C = 11$.
For digits, it's common to use overline: $\overline{abc}_k = ak^2 + bk + c$
A: From the second equation, so $c = A^2$. Apply the $C$ value in the first equation, so $B = 1$. 
So $A + B + C = (A \cdot A) + A + 1$. If $A = 3$, then $(3 \cdot 3 ) + 3 + 1 = 13$
A: We have the equations
$$A^2-10B-C=0$$
$$A^3-10C-A=0$$
that  lab bhattacharjee established.
also we know that
$$\prod_{i=0}^9(A-i)=0$$
because $A \in \{0,1,2,3,4,5,6,7,8,9\}$.
Also we have 
$$\prod_{i=0}^9(B-i)=0$$
$$\prod_{i=0}^9(C-i)=0$$
Now I calculate the grobner basis ${6 C-C^2,3 A-2 C,C-6 B}$    of this system using Maxima
(%i1) load(grobner)$
(%i2) poly_reduced_grobner([A^2-10*B-C,A^3-10*C-A,prod(A-i,i,0,9),prod(B-i,i,0,9),prod(C-i,i,0,9)],[A,B,C]);
(%o2) [6*C-C^2,3*A-2*C,C-6*B]

Solving  the system of equations
$$
\begin{eqnarray}
6C-C^2&=&0  \\
3A-2C&=&0 \\
C-6B&=&0
\end{eqnarray}
$$
we get two solution triples$(A,B,C)$:  $(0,0,0)$ and $(4,1,6)$
