When you have factored the expression completely, try evaluating it for some positive integers to see if it works. Why is that the case?
See solution.
Practice makes perfect
First, notice that in the given expression both terms have n as common factor. We can rewrite it as shown below.
n^3-n &= (n)n^2-(n)1
&= n (n^2-1 )
Now, since 1=1^2, we can see that the expression in parentheses is a difference of two squares, which we know how to factor.
x^2 - y^2 = ( x+ y)( x- y)
By following this pattern, we can factor the given expression completely.
n^3-n &= (n)n^2-(n)1
&= n( n^2- 1)
&= n( n+ 1)( n- 1)
Using the Commutative Property of Multiplication, we can rewrite the given expression as n^3-n= (n-1)(n)(n+1). We can now evaluate it for some positive integer values of n, and see if the obtained result is a multiple of 3.
n
(n-1)(n)(n+1)
Simplifying
Multiple of 3?
1
( 1-1)( 1)( 1+1) = 0
(0)(1)(2)=0
3*0=0 âś“
2
( 2-1)( 2)( 2+1)
(1)(2)( 3)=6
3* 2= 6 âś“
3
( 3-1)( 3)( 3+1)
(2)( 3)(4)= 24
3* 8= 24 âś“
4
( 4-1)( 4)( 4+1)
( 3)(4)(5)= 60
3* 20= 60 âś“
5
( 5-1)( 5)( 5+1)
(4)(5)( 6)= 120
3* 40= 120 âś“
6
( 6-1)( 6)( 6+1)
(5)( 6)(7)= 210
3* 70= 210 âś“
7
( 7-1)( 7)( 7+1)
( 6)(7)(8)= 336
3* 112= 336 âś“
8
( 8-1)( 8)( 8+1)
(7)(8)( 9)= 504
3* 168= 504 âś“
The statement seems to be true. Why is that so? Notice that (n-1)(n)(n+1) are three consecutive numbers. No matter what number we use, one of them will always be a multiple of 3. Therefore, the product (n-1)(n)(n+1)= n^3-1 will be a multiple of 3 as well, for any positive integer n.
