How extra are you?

We consider the size of $m$ relative to $n$ . If $m = n$ , then $n$ is a extra number if and only if $7n = 133$ . In this case, we get $19$ is an extra number.

Now, suppose that $m \geq n + 1$ . Then

$n^3 + 7n - 133 = m^3 \geq (n + 1)^3 = n^3 + 3n^2 + 3n + 1$

so that $\displaystyle 3n^2 - 4n + 134 \leq 0$ . For positive integers $n$ , it is easy to see that this is impossible.

It remains to consider $\displaystyle m \leq n - 1$ . For such $m,$ we have

$\displaystyle n^3 + 7n - 133 = m^3 \leq (n - 1)^3 = n^3 - 3n^2 + 3n - 1$

which implies $\displaystyle 3n^2 + 4n - 132 \leq 0$ . It follows here that $n \leq 6$ . One checks that

$\displaystyle 6^3 + 7 \cdot 6 - 133 = 5^3$

and

$5^3 + 7 \cdot 5 - 133 = 3^3$ ,

so $6$ and $5$ are extra numbers. For $n \leq 4$ , we see that $\displaystyle n^3 + 7n - 133 < 0$ and hence, $\displaystyle n^3 + 7n - 133$ cannot equal $m^3$ for a positive integer $m$ .

Therefore, the sum of all extra numbers is $19 + 6 + 5 = 30$ .