00:01
Any integer is of the form any integer is of the form 3k 3k plus 1 3k plus 2 now if you take 3k the cube of 3k 3k whole cube is 27 k cube which i can write it as 9 into 3k cube now 3k cube is of the form 9p where p is 3k cube and p is of obviously an integer.
00:37
Now if you take k plus one whole cube, let's use a plus b whole cube formula it is a cube plus b cube plus 3ab that is 3 into 3k into 1 into a plus b.
00:51
Now obviously this will be a nine so nine times an integer call it as nine r and this is obviously nine times some integer 9p now sum of 9p in nine r is some nine cube plus one cube is 1 so it is of the form 9q plus 1 that means cube of this is of the form 9p in the cube of this is of the form 9 q plus 1 now let's take the third case 3 k plus 2 whole cube alright so now 3k plus 2 whole cube so we have 3k plus 2 whole cube let's use a plus b b whole cube we have a cube plus b plus b that is 2 a b that is 3 into 3k into 2 into a plus b that is a tk plus two now this is again nine so nine r and this is obviously of nine p so it will be nine p plus nine r is some nine m plus two cube is eight so final conclusion is the cube of any integer is eight of the form nine p or nine p plus one as we have seen in the last place and third case nine p plus eight so only in these three formats the cube of integer will be.
02:08
Now let us prove that this is always an integer for any national number n, but we should not use mathematical induction, right? so let's see how can it be done without using mathematical induction.
02:24
Let me, uh, let's take one lemma.
02:27
Let's take one lemma.
02:29
So the lemma is the product of the consecutive national numbers, the product of the product three consecutive consecutive natural numbers is always divisible by is always divisible by six so let's prove this lemma so now let's take a natural number n and it's sucks n plus one and the next one it is n is n plus two this thing we need to prove that it is always divisible by six let's call ff n as n into n plus 1 into n plus 2 now any national number n is of the form so let's take case 1 n is 3m there is a multiple of 3 so let's plug in in f of n so we have 3m into 3m plus 1 into 3m plus 1 into 3m plus 2 right so now that can be written as 3m into multiply these two you get 9m square plus 9m plus 2 so multiply further 27 m cube plus 27 m squared plus 6m now which can be further written as 24m cube plus 34 m square plus 6m plus 3m cube plus 3m squared plus 3m square but this guy is obviously a multiple of 6 but how about this guy this guy can write it as 3m into m plus 1 but if m is even 3 into even number is multiple of 6 and if m is odd in case then m plus 1 will be even and 3 into even multiple of 6 so in any case this will be obviously multiple of 6 so the totality is the multiple of 6 so for n is 3m we have 2 now let's jump on to the case 2 so what is case 2 case 2 is basically n is of the form 3m plus 1 all right so let's take that now when it is of the form 3m plus 1 your f of n will be 3m plus 1 into 3m plus 2 into 3m plus 3m all right so now that can be written as this guy can read it as 3 times of m plus 1 into now let's multiply these two these two when you multiply you will have 9m square plus 9m plus 2 all right so now that can be written as 3 times of m plus 1 into 9m squared plus 9m equal reader is 9m into m plus 1 and plus this t times m plus 1 multiply with 2 that is 6 times of m plus 1...