Show that the difference between any integer and its cube is always divisible by 6

Question

Show that the difference between any integer and its cube is always divisible by 6

Answer 1

$$\begin{aligned} \text{Diff}&=x^3-x \\ &=x(x^2-1) \\ &=x(x+1)(x-1) \qquad \text{via }(a^2-b^2)=(a+b)(a-b) \end{aligned}$$

$=(x-1)\cdot x\cdot (x+1)$

So that always turns out to be product of three consecutive integers...which will always have one even number as 2*m and one number divisible by 3 as 3*n. Hence the product will always have minimal factors as 2 and 3...So, it is divisible by 6. That's it!!!

Answer 2

$$ x^3-x=(x-1)\cdot x\cdot (x+1). $$ Between three consecutive numbers, at least one is even, and at least one is multiple of $3$.

Answer 3

$$x-x^3=x(1-x^2)=x(1-x)(1+x)=-(x-1)x(x+1)$$ since $x-1,x,x+1$ are three consecutive numbers and their product is divisible by 6.

Answer 4

$$n^3-n=(n+1) n (n-1)= 6\binom{n+1}{3}$$

Answer 5

For the six first $n$, we get $n^3-n=0,0,6,24,60,120$.

Then $(n+6k)^3-(n+6k)=n^3-n+6m$.

Answer 6

If you factorize $n^3 − n$ you will get $(n-1)\cdot n\cdot (n+1)$ and you should know that product of $m$ consecutive integers is divisible by $m!$. (See here The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients))

Since $(n-1)\cdot n\cdot (n+1)$ is product of $3$ consecutive integers thus it will be divisible by $3!=6$.