If you can't think of that number, you can break down Q into
its prime factorization and look for equal groups of numbers.
Let's draw a factor tree.
So the prime factorization of Q is PRIMES.join( "\\times " ).
\Large{\sqrt[3]{Q} = \text{?}}
N
\sqrt[3]{Q} is the number that, when
multiplied by itself three times, equals Q.
We're looking for \sqrt[3]{Q}, so we want to split the prime factors into three identical groups.
We only have three prime factors, and we want to split them into three groups, so this is easy.
Q = PRIMES.join( "\\times " ), so N^3 = Q.
Notice that we can rearrange the factors like so:
Q = PRIMES.join( "\\times " ) = \left([ F_N.join( "\\times " ), F_N.join( "\\times " ), F_N.join( "\\times ") ].join( "\\right)\\times\\left(" )\right)
So \left(F_N.join( "\\times " )\right)^3 = N^3 = Q.
So N^3 = Q.
So \sqrt[3]{Q} is N.