The best thing is to do a prime factorization of this number and look for 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 we found that the prime factorization of

is `Q`

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

What is the square root of

?`Q`

`N`

We're looking for the square root of

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

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

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

.`N`^2 = `Q`

Notice that we can rearrange the factors like so:

`Q` = `PRIMES.join(" × ")` = \left(`F_N.join( "\\times " )`\right) \times \left(`F_N.join(" × ")`\right)

So `\left(`

.
`F_N.join( "\\times " )`\right)^2 = `N`^2 = `Q`

So

.
`N`^2 = `Q`

So the square root of

is `Q`

.`N`

What is the cube root of

?`Q`

`N`

We're looking for the cube root of

, 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 the cube root of

is `Q`

.`N`