To find all 3digit numbers $abc$ (in base $10$) such that
$$ a + b^2 + c^3 = 100a + 10b + c $$ Rearranging gives $$c^3  c  99a = b (10  b)$$ $$c(c+1)(c1)  99a = b (10  b)$$
For any three consecutive integers one of them is divisible by
$3$. Since $3$ divides $99$ it follows that $3$ divides $b(10b)$.
Since $3$ is a prime this limits the possible choices of $b$:
Either
$b=0 10b=10$  $b(10b) = 0$  
$b=3 10b=7$  or $b=7 10b=3$  $b(10b) = 21$ 
$b=6 10b=4$  or $b=4 10b=6$  $b(10b) = 24$ 
$b=9 10b=1$  or $b=1 10b=9$  $b(10b) = 9$ 
Hence the possible values of $b(10b)$ are $0$, $9$, $21$ and $24$.
We now have to find a multiple of $99$ which when subtracted
from a product of $3$ consecutive natural numbers gives $0$, $9$,
$21$ or $24$.
Since $a$ is at least $1$ $c(c+1)(c1)$ is at least $99$ so $c$ is
at least $5$.
$c$  $c(c+1)(c1)$  $a$  $99a$  $c(c+1)(c1)99a$ 

$5$  $120$  $1$  $99$  $21$ 
$6$  $210$  $2$  $198$ 
$12$

$7$  $336$  $3$  $297$  $39$ 
$8$  $504$  $5$  $495$  $9$ 
$9$  $720$  $7$  $693$  $27$ 
(Since $0$, $9$, $21$, $24 < 99$ only the multiple of $99$ which is closest to $(c+1)c(c1)$ needs to be checked.)
From the table we can see that the following are the possibilities for $a$, $b$ and $c$:
$a=1$ $b=3$ or $b=7$ $c=5$
$a=5$ $b=1$ or $b=9$ $c=8$
giving the solutions $135$, $175$, $518$ and $598$.