The best thing is to do a prime factorization of this number and look for groups of numbers.
Let's draw a factor tree.
So we found that the prime factorization of Q
is PRIMES.join( "\\times " )
.
What is the square root of Q
?
N
We're looking for the square root of Q
, so we want to split the prime factors into two identical groups.
We only have two prime factors, and we want to split them into two groups, so this is easy.
Q = PRIMES.join( "\\times " )
, so 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 Q
is N
.
What is the cube root of Q
?
N
We're looking for the cube root of 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 the cube root of Q
is N
.