Neither All? What's Left?

for each integer n from 1 to 1000 there is a number $n^2$ that is a perfect square less than or equal to 1,000,000 so there are 1000 perfect squares

for each integer m from 1 to 100 there is a number $m^3$ that is a perfect cube that is equal to or less than 1,000,000 so there are 100 perfect cubes

how many of these perfect cubes are also perfect squares?

integers raised to the sixth power will be both perfect squares and perfect cubes; there are ten of these $(1^6, 2^6, 3^6, ...10^6)$ that produce numbers in this range

so there are 1000 + 100 - 10 = 1090 separate numbers that are either perfect squares or cubes in the given range

meaning that 1,000,000 - 1090 = 998,910 are neither perfect squares or cubes

Python 2.7:

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perfect = {2:set(),3:set(),4:set()}
for power in xrange(2,5):
    new = 0
    while new**power < 1000000:
        perfect[power].add(new**power)
        new += 1
count = 10**6 - len(perfect[2] | perfect[3] | perfect[4])
print "Answer:", sum([int(c) for c in str(count)])

Number of perfect squares = $1000$ Since we know that ${1000}^{2} = {10}^{6}$

Number of perfect cubes = $100$ Since we know that ${100}^{3} = {10}^{6}$ But we have already counted $1$ hence we have $99$

Number of perfect fourth powers = $31$ Since ${31}^{4}$ is the closest to ${10}^{6}$ But we have already counted these fourth powers in the squares. Hence we have $0$

Total = $1000 + 99 + 0 = 1099$

${10}^{6} - 1099 = \boxed {998901}$

$9 + 9 + 8 + 9 + 1 = \boxed {\huge 36 }$