6. Remove from the set
$$
A=\left\{n \mid n \leqslant 10000, n \in \mathbf{Z}_{+}\right\}
$$

all perfect squares and perfect cubes, then arrange the remaining elements in ascending order to form a sequence. The 2014th term of this sequence is $\qquad$
6.2068.

Among $1,2, \cdots, 2070$, since
there are 45 perfect squares (because $45^{2}=2025$, $46^{2}=2116$), 12 perfect cubes (because $12^{3}$ $=1728,13^{3}=2197$), and 3 perfect sixth powers (because $3^{6}=729,4^{6}=4096$), and by
$$
2070-45-12+3=2016 \text {, }
$$

we know that 2070 is the 2016th term of the sequence.
Since $2070, 2069, 2068$ are neither perfect squares nor perfect cubes, thus, the 2014th term of this sequence is 2068.