Yes
No

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1^3=1 , 9^3=729

10^3=1000 , 99^3=970299

100^3=1000000 , 999^3 = 997002999

let the number of digits of a = n then the number of digits of a^3 = 3n-2 or 3n-1 or 3n

number of digits of a + number of digits of a^3 =4n-2 or 4n-1 or 4n

therefore n=2017/4 or n=2018/4 or n=2019/4 none of which is valid because n must be an integer