<s>[INST] A question about how to factorise a binomial. If $x=\frac{-1}{3}$ 
$(x+1)^3-{3(x+1)}^2+3(x+1)$ =?
Using the binomial theorem coefficients I'm going to add 1 at the end 
[$(a-b)^3=a^3-3a^2b+3ab^2-b^3]
So we now have 
$(x+1)^3-{3(x+1)}^2+3(x+1)+1$ 
How do I continue from here? 
 [/INST] Note that by
$$(a-b)^3=a^3-3a^2b+3ab^2-b^3$$
with


*

*$a=x+1$

*$b=-1$


we have
$$(x+1)^3-{3(x+1)}^2+3(x+1)=(x+1)^3-{3(x+1)}^2+3(x+1)-1+1=[(x+1)-1)]^3+1=x^3+1$$
</s>