Three digit number $ABC$ with $ABC = A + B^2 + C^3$ Is there a trick for solving this problem about number of digits?

$ABC$ is a three-digit natural number, such that $ABC = A + B^2 + C^3$.
According to above equation what is $ABC$ ?

 A: The following is not really a trick, but at least it's a computation which can be done by hand.
First note that $10B\ge B^2$ (since $0\le B\le9$). Hence 
$$100A+10B+C=A+B^2+C^3\implies 99A+C<C^3$$
and since $A\ge1$, this gives us immediately that $C\ge5$. 
Now, assume that $B=0$. Then we have $100A+C=A+C^3$, but since none of $5^3=125$, $6^3=216$, $7^3=343$, $8^3=512$, and $9^3=729$ are in the range $[100n-10,100n+10]$, there are no solutions with $B=0$, so from here on we assume $B\ne 0$
Now we have $10B>B^2+A$, so $100A<C^3$, and thus $A$ is at most the first digit of $C^3$, but since we also know $A$ must be at least the first digit of $C^3$ (easy to see from the original formulation of the problem), we have that $A$ is exactly the first digit of $C^3$.
Now all we have to do is solve $5$ quadratic equations for $B$ (one for each possible value of $C$) and see if there are integer solutions:
$C=5$:$$100+10B+5=1+B^2+125\implies B=3,7$$
$C=6$:$$200+10B+6=2+B^2+216\implies B=5\pm\sqrt{13}$$
$C=7$:$$300+10B+7=3+B^2+343\implies B=5\pm i\sqrt{14}$$
$C=8$:$$500+10B+8=5+B^2+512\implies B=1,9$$
$C=9$:$$700+10B+9=7+B^2+729\implies B=5\pm i\sqrt{2}$$
At long last this gives us the $4$ solutions $ABC\in\{135,175,518,598\}$.
