(n x n x n -n) is divisible by 3.explain the reason
Answers
= n(n² - 1)
= n(n² - 1²)
= n[(n + 1)(n - 1)]
= (n - 1)(n)(n + 1)
We know that every positive integer is of the form 3q, 3q + 1 or 3q + 2, for some integer 'q'.
When n = 3q
(n - 1)(n)(n + 1) = (3q - 1)(3q)(3q + 1)
n³ - n = (3q - 1)(3q)(3q + 1)
n³ - n is divisible by 3 ... (i)
When n = 3q + 1
(n-1)(n)(n+1)=(3q + 1 - 1)(3q + 1)(3q + 1 + 1)
n³ - n = (3q)(3q + 1)(3q + 2)
n³ - n is divisible by 3 ... (ii)
When n = 3q + 2
(n-1)(n)(n+1)=(3q + 2 - 1)(3q + 2)(3q + 2 + 1)
n³ - n = (3q + 1)(3q + 2)(3q + 3)
n³ - n = (3q + 1)(3q + 2)[3(q + 1)]
n³ - n = 3(3q + 1)(3q + 2)(q + 1)
n³ - n is divisible by 3 ... (iii)
From (i), (ii) and (iii)
we see that, for every possible form of a positive integer 'n', n³ - n is divisible by 3.
... Hence Proved!
*mark me the brainliest!*
Question:
Explain the reason why n^3 - n is divisible by 3.
Explanation:
(1) Mathematical induction...
Let n = 1
n^3 - n = 1^3 - 1 = 1 - 1 = 0
As 0 is divisible by any integer, n^3 - n is divisible by 3 if n = 1.
Let n = 2.
n^3 - n = 2^3 - 2 = 8 - 2 = 6
As 6 is divisible by 3, then so is n^3 - n, if n = 2.
Let n = 3.
n^3 - n = 3^3 - 3 = 27 - 3 = 24
As 24 is divisible by 3, then so is n^3 - n, if n = 3.
So, let n = k.
Assume that k^3 - k is divisible by 3.
Thus, let n = k + 1.
=> n^3 - n
=> (k + 1)^3 - (k + 1)
=> k^3 + 3k^2 + 3k + 1 - k - 1
=> k^3 - k + 3k^2 + 3k
=> k^3 - k + 3k(k + 1)
Here, we have already assumed earlier that k^3 - k is divisible by 3. To this, 3k(k + 1), which is also divisible by 3, is added.
Thus, if k^3 - k is divisible by 3, then so will be (k + 1)^3 - (k + 1).
Hence n^3 - n is always divisible by 3, for any positive integer 3.
(2) Factorization (General Method)
Let n^3 - n be factorized...
=> n^3 - n
=> n(n^2 - 1)
=> n(n - 1)(n + 1)
From here, we can find and say that n^3 - n is the product of any three consecutive integers, where n is the mean and median of these three consecutive integers.
When we consider any three consecutive integers, it is always and absolutely true that one among the three is divisible by 3. It can be found out by examples. Or, let's have a look at the given below:
Suppose n is a multiple of 3, and can be written as 3k.
Thus,
=> n^3 - n
=> (3k)^3 - (3k)
=> 27k^3 - 3k
=> 3k(9k^2 - 1)
OR
=> n(n - 1)(n + 1)
=> 3k(3k - 1)(3k + 1)
=> 3k((3k)^2 - 1^2)
=> 3k(9k^2 - 1)
Here it seems that n^3 - n is a multiple of 3, or 3k, can be. Thus proved!
Let n - 1 be a multiple of 3, and be 3k. So that n becomes 3k + 1.
=> n^3 - n
=> (3k+ 1)^3 - (3k + 1)
=> 27k^3 + 27k^2 + 9k + 1 - 3k - 1
=> 27k^3 + 27k^2 + 6k
=> 3k(9k^2 + 9k + 2)
OR
=> n(n - 1)(n + 1)
=> (3k + 1)(3k)(3k + 2)
=> 3k((3k)^2 + 3(3k) + 2)
=> 3k(9k^2 + 9k + 2)
Here it also seems...!!!
Let n + 1 be multiple of 3, and be 3k. Thus n becomes 3k - 1.
=> n^3 - n
=> (3k - 1)^3 - (3k - 1)
=> 27k^3 - 27k^2 + 9k - 1 - 3k + 1
=> 27k^3 - 27k^2 + 6k
=> 3k(9k^2 - 9k + 2)
OR
=> n(n - 1)(n + 1)
=> (3k - 1)(3k - 2)(3k)
=> 3k((3k)^2 - 3(3k) + 2)
=> 3k(9k^2 - 9k + 2)
Here it too...!!!
Thus proved!!!