By definition, the cubic root of a number #x# is a number #y# such that #y^3=x#.
Apart from using the calculator, of course, you can see if a number #n# is a perfect square by factoring it into primes, and if the number has a representation of the form
#n= p_1^{d_1} \times p_2^{d_2}\times ... \times p_n^{d_n}#, then it is a perfect cube if and only if every #d_i# is divisible by 3.
Factoring #128# in primes gives you
#128=2^7#, thus it is not a perfect cube (i.e., its cube root is not an integer).
Anyway, we can say that the cubic root of #128# is #128# to the power of #1/3#, so we have
#128^{1/3}=(2^7)^{1/3}=2^{7/3}=2^{2+1/3}#
By using the formula #a^{b+c}=a^b\cdot a^c#, we have that
#2^{2+1/3}=2^2 \cdot 2^{1/3}=4 \cdot 2^{1/3}#
which is four times the cubic root of #2#
