A question about how to factorise a binomial.

Question

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?

Answer 1

Note that by

$$(a-b)^3=a^3-3a^2b+3ab^2-b^3$$

with

• $a=x+1$
• $b=-1$

we have

$$(x+1)^3-{3(x+1)}^2+3(x+1)=(x+1)^3-{3(x+1)}^2+3(x+1)-1+1=[(x+1)-1)]^3+1=x^3+1$$