pretty confusing math problem help please?

Question

The sequence 2, 3, 5, 6, 7, 10, 11, ... contains all the positive integers from least to greatest that are neither squares nor cubes. What is the 400th term of the sequence?

Answer 1

From 2 to 401, we have 400 numbers.
Out of this, we have to remove 19 squares (2 to 20) and 6 cubes (2 to 7), but add 1 (64) which is both a square and a cube. That pushes us from 401 to 401+19+6-1=425.
Between 401 and 425, there are neither cubes (next=512) nor squares (next=441).
So the 400th term is 425.

Answer 2

Consider the number 401. This number has floor(√401) = 20 squares less than it, and floor(∛401) = 7 cubes less than it, but 2 of these numbers (specifically, 1 and 64) are both squares and cubes, there are 20+7-2 or 25 numbers less than 401 which aren't in the sequence, which means that there are 400-25=375 numbers in the sequence before it. So 401 is the 376th number in the sequence. All of the numbers from 402-425 are also in the sequence (the next number not in the sequence is 21² or 441), and there are 24 such numbers, thus 425 is the 400th term in the sequence.

Answer 3

424.

Answer 4

640?