Australian Mathematics Competition Junior last question

Regroup terms to get $99a + b(10-b) = c^3-c$ . Note that $b(10-b)$ is at most $25$ .

Now if $a \ge 6$ then $c^3-c \ge 594$ , which is only possible if $c = 9$ . But if $99a + b(10-b) = 720$ , then $a \le 7$ and $b(10-b) \ge 720-(99 \cdot 7) = 27$ , which is no good.

So let's try $a=5$ . This gives $495 + b(10-b) = c^3-c$ . Clearly we must have $c=8$ , whence $b(10-b) = 9$ . This has two integer solutions, and we take the larger one, $b = 9$ . So the solution is $\fbox{598}$ .

a + b² + c³ = 100a + 10b + c
c³ – c = (100a + 10b) – (a + b²)
c(c² – 1) = a(100 – 1) + (10b – b²)
(c – 1)(c)(c + 1) = 99a + b(10 – b)

LHS is the product of 3 consecutive numbers with c < 10, and is always divisible by 6, while RHS's first term is already a multiple of 3 but the second is not. So b ≠ { 2, 5, 8 } and both a and b from RHS must be of the same parity.

Possible numbers are 135, 175, 518 & 598.