The tricky part is to evaluate the decimals, but we can keep multiplying it by 10 until we have the first 10 digits. To check if all numbers are unique, throw everything into a set, which by definition doesn't contain duplicates, and the length should be 10.

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def very_special(x):
    y = pow(x, 1.0 / 3)
    z = str(int(y * pow(10, 10)))

    digits = []
    for i in range(10):
        digits.append(z[i])

    digits = set(digits)
    return len(digits) == 10

num = 1
while True:
    if very_special(num):
        print num
        break
    num += 1

A shorter solution in python, could be:

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"A Very Special Number"

n = 1
while len(set( "{0}".format(n**(1/3)).replace('.','')[0:10])) != 10:
    n += 1
print(n)

Output

2017

Basically, I start with $n=2$ and I evaluate $\sqrt[3]{n}$ for each loop. Now I take each digit $d_i\in[0,1,...,9]$ , $1 \leq i \leq 10$ and I put it in an empty array $A$ such that $A_{d_i}=d_i$ . For example, if I'm looking at the $5$ th digit of the root whose value is $6$ ( $d_5=6$ ), I'll put it in $A_6$ . If $d_i=0$ , than $A_{10}=0$ . If I'm looking at a digit for the second time, its place in $A$ will be already occupied and than the loop stops and goes for $n+1$ . The algorithm finds a solution when $A=[1,2,3,...,0]$ .

MATLAB code: