0 0 randRange( 2, 12 ) getPrimeFactorization( N )

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.

init({ range: [ [-1, FACTORIZATION.length + 2], [ -2 * FACTORIZATION.length - 1, 1] ], scale: [30, 30] }); label( [cx + 1, y], curr );
path( [ [cx + 1, y - 0.5], [cx, y - 1.5] ] ); path( [ [cx + 1, y - 0.5], [cx + 2, y - 1.5] ] ); y -= 2; cx += 1; curr = curr / factor; label( [cx - 1, y], factor ); circle( [cx - 1, y], 0.5); label( [cx + 1, y], curr );
circle( [cx + 1, y], 0.5);

So the prime factorization of `Q` is `PRIMES.join( "\\times " )`.

N * N * N getPrimeFactorization( Q ) PRIMES.slice( 0, PRIMES.length - 1 ) Q

`\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`.