6. The number of positive integers less than or equal to $10^{6}$ that are neither perfect squares, nor cubes, nor perfect fourth powers is $\qquad$ .
6. 998910 .

There are $\left[10^{3}\right]=1000$ perfect squares, $\left[10^{2}\right]=100$ cubes, and $\left[10^{\frac{3}{2}}\right]=31$ fourth powers. There are 10 numbers that are both perfect squares and cubes, 31 numbers that are both perfect squares and fourth powers, and 12 numbers that are both cubes and fourth powers. Since $(\sqrt{10})^{12}=10^{6}$, there are $(\sqrt{10})=3$ numbers that are 12th powers and less than $10^{6}$, which means they are all three. Therefore, the answer is $10^{6}-$ $1000-100-31+10+31+3-3=998910$ (numbers).