Find out the missing term of the number series given below.

68, ?, 407, 470, 496, 503

The correct answer is 283.

To find the missing term of the number series, let us analyze the differences between consecutive terms starting from the end of the series where all numbers are given:

First, let's find the difference between 496 and 503:
$503-496=7$

Next, find the difference between 470 and 496:
$496-470=26$

Next, find the difference between 407 and 470:
$470-407=63$

Let's analyze the pattern of these differences:
The differences are 63, 26, and 7.
Notice that these numbers are close to perfect cubes ($n^3$):
$63=4^3-1$ (since $64-1=63$)
$26=3^3-1$ (since $27-1=26$)
$7=2^3-1$ (since $8-1=7$)

Following this pattern, the differences between consecutive terms in the series follow the form $n^3-1$ decreasing by 1 for each step. Let's write out the full sequence of differences:
Between 1st and 2nd term: $6^3-1=216-1=215$
Between 2nd and 3rd term: $5^3-1=125-1=124$
Between 3rd and 4th term: $4^3-1=64-1=63$
Between 4th and 5th term: $3^3-1=27-1=26$
Between 5th and 6th term: $2^3-1=8-1=7$

Now, let's verify by calculating the missing term (let it be $x$):
Using the first difference ($215$):
$x=68+215=283$

Let's double-check using the next difference ($124$):
$283+124=407$

This matches the third term in the series (407). Therefore, the missing term is indeed 283.