SOLUTION: What is the smallest number by which 26244 must be divided to give a perfect cube? Also find the cube root of the quotient.

                    
        26244
         / \
        2  13122
           / \
          2  6561
             / \
            3  2187
               / \
              3  729
                 / \
                3  243
                   / \
                  3  81
                     / \
                    3  27
                       / \
                      3   9
                         / \
                        3   3
                      


So   26244 = 2238 

To become a cube, all the prime factors of it must 
be to a power which is a multiple of 3.

Notice that the prime number base 2 in the factorization
is raised to the 2nd power (exponent), but exponent 2 is 
NOT a multiple of 3, so we'll have to multiply by the 1st 
power of 2 so that when we add exponents of 2 we will get 23.

Notice also that the prime number base 3 in the factorization
is raised to the 8th power (exponent), but exponent 8 is 
NOT a multiple of 3, so we'll have to multiply by the 1st 
power of 3 so that when we add exponents of 3 we will get 39.

So we have to multiply by 2131
or 2∙3 or 6 to cause 26244 to become a perfect cube.

So we have to multiply 2238 by 2131
so that it will become 2339 and both prime bases 2 and 3
will be raised to powers (exponents) which are both multiples of 3.

So then the cube root of 2339 will be gotten
by dividing each exponent by 3, which will give 2133 which
is 2∙27 or 54.

That's the same as saying

The 26244 must be multiplied by 6 gives 157464 which is a 
perfect cube.  It is a perfect cube because 543 = 157464.

And the cube root is 54 because 54∙54∙54 = 157464.

Edwin