If you can't think of that number, you can break down

into
its prime factorization and look for equal groups of numbers.
`Q`

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

is `Q`

.`PRIMES.join( "\\times " )`

What is `\sqrt[3]{`

?`Q`}

`N`

`\sqrt[3]{`

is the number that, when
multiplied by itself three times, equals `Q`}

.
`Q`

We're looking for `\sqrt[3]{`

, so we want to split the prime factors into three identical groups.`Q`}

We only have three prime factors, and we want to split them into three groups, so this is easy.

, so `Q` = `PRIMES.join( "\\times " )`

.`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]{`

is `Q`}

.`N`