What is the smallest number by which \[18522\] must be divided so that the quotient will is a perfect cube?

Hint: The prime factorization is a method which is used to express a given number in form of multiplication of prime numbers. On the other hand, in factorization if a prime number occurs more than once, then factorization is expressed in exponential form.
To solve such type of questions i.e quotient will be a perfect cube or perfect square, we always do the prime factorisation of the given number and make the groups in triplet (to calculate perfect cube) and doublet (for perfect square). Here triplet means group of three same numbers and doublet means group of two same numbers.

Complete step by step solution: We will start here with prime factorisation of our given number:


So Prime factorisation of \[18522{\text{ }} = {\text{ }}2 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7\]
Since we have to make quotient a perfect cube, hence we make group of three same number here.
i.e. \[18522{\text{ }} = {\text{ }}2 \times \]\[3 \times 3 \times 3\]\[ \times \]\[7 \times 7 \times 7\]
Clearly it can be seen that 2 is the only factor which is extra here. If we divide our number by 2 here then its quotient can become a perfect cube here.
i.e. \[\dfrac{{18522}}{2} = 9261\] which is a perfect cube of \[21\].

Hence, our desired answer is 2.

Note: Students always remember in prime factorisation you have to use prime numbers. Prime numbers are those which can be divided by one or the number itself. Example \[2,3,5,7 \ldots .\]
Also one is not a prime number. Many students use numbers other than prime and get wrong answers. Secondly, always remember you have to choose the same number to make a triplet because \[{2^3} = 2 \times 2 \times 2\], the cube is always a group of three numbers. We divide only with that number which is not forming a triplet.