I know this is a really basic question for most of you, but I just can't figure it out.
I am currently learning about induction in university and I have a question about the following:
$$ ((n^3+3n^2+3n+1)-(n+1))\,\,mod\,\,3 = 0 $$ transforms into: $$ ((n^3-n)+(3n^2+3n))\,\,mod\,\,3 = 0 $$
Now I know that the first one is a binomial formula of $(n+1)^3$, but I can't figure out how to get from the first to the second line
Since multiplication is distributive over addition, for any real numbers $a$, $x$, and $y$ we have $$ a(x+y)=ax+ay \, . $$ For the special case $a=-1$, $$ -(x+y)=-1(x+y)=-1x+(-1)y=-x-y \, . $$ Hence, \begin{align} n^3+3n^2+3n+1-(n+1) &= n^3+3n^2+3n+1-n-1 \\ &= n^3+3n^2+2n \end{align} This is the same as $(n^3-n)+(3n^2+3n)$.