What is the fastest way to calculate $(x+1)^3-(x-1)^3$? 
What is the fastest way to calculate $(x+1)^3-(x-1)^3$?

My teacher gives me this question, and I use 5 minutes to calculate it, and there is an error. I want to know if there is a faster method to do it.
 A: I like Pascal's triangle:   the third row gives the coefficients of $(x+a)^3$, namely,  $1331$.
So we get $(x+1)^3=x^3+3x^2+3x+1$ and $(x-1)^3=x^3-3x^2+3x-1$.
Thus the difference is $6x^2+2$.
A: $$(1+x)^3+(1-x)^3$$ is an even polynomial of degree at most three. Hence the expansion has only even terms and must be of the form
$$ax^2+b.$$
Setting $x=0$, you find $b=2$, then with $x=1$, $a+b=8$ and you are done.

Alternatively, you may happen to know by heart the fourth row of Pascal's Triangle, $1\ 3\ 3\ 1$, and you get the coefficients $1+1,3-3,3+3,1-1$ (by increasing powers).
A: Use the fact that$$a^3-b^3=(a-b)(a^2+ab+b^2).$$
A: $\mathbf{Method\ 1}:$ $$S=(x+1)^3 - (x-1)^3 = 2((x+1)^2 + (x-1)^2 + (x+1)(x-1))$$$$=2(2(x^2+1) + x^2 -1)$$ $$=2(3x^2 + 1)$$ $$=6x^2+2$$ $\mathbf{Method\ 2:}$ $$(x+1)^3 = \displaystyle\sum_{r=0}^3\binom 3rx^r$$ $$(x-1)^3=\displaystyle\sum_{r=0}^3(-1)^r\binom 3rx^r$$ Clearly in their addition only terms with even $r$ remain. Here, they are $r = 0,2$. So the sum is$$2\left(\binom 30 + \binom 32x^2\right)$$ $$=6x^2 + 2$$ In any case we have the solution $$\boxed{S = 6x^2+2}$$
A: Mostly it depends how fast you in calculation. It's same formula but different ways to write
$$(a+b)^3=a^3+b^3+3ab(a+b)$$
$$(x+1)^3=x^3+1+3x(x+1)$$
$$(x-1)^3=x^3-1-3x(x-1)$$
$$... - .. +..+ $$
$$S=2+3x(x+1+x-1)=6x^2+2$$
