 Hi and welcome to the session. Let us discuss the following question. Question says, find the smallest number by which 128 must be divided to obtain a perfect tube. Let us now start with the solution. Now prime factorization of 128 is 2 multiplied by 2 multiplied by 2 multiplied by 2 multiplied by 2 multiplied by 2 multiplied by 2 or you can say multiplying 2 7 times we get 128. Now understand that in the prime factorization of any number is each factor appears 3 times then the number is a perfect cube. Now clearly you can see here we have grouped the factors in triplets. Now only 1 2 remains after grouping the 2's in triplets. So 128 is not a perfect cube. So we can write 128 is not a perfect cube. Now observe that if we divide 128 by 2 then prime factorization of quotient that is 128 divided by 2 will not contain this 2. So we can write if we divide 128 by 2 then prime factorization of quotient does not contain remaining 2. Now on dividing both the sides of this expression by 2 we get 128 divided by 2 is equal to 2 multiplied by 2 multiplied by 2 multiplied by 2 multiplied by 2 multiplied by 2 multiplied by 2. Now this is the quotient and this is the prime factorization of this quotient. Now clearly you can see nothing remains after grouping the factors in triplets. So quotient is a perfect cube. So we get the smallest number by which 128 should be divided to make it a perfect cube is 2. And the resulting perfect cube is 128 divided by 2 which is equal to 64 and 64 is a perfect cube. Or we can say 64 is equal to 4 cube. So our required answer is 2. This completes the session. Hope you understood the solution. Take care and bye for now.