Question 332064
{{{(n^2-1)/(n+1)^2}}} Start with the given expression.



{{{((n+1)(n-1))/(n+1)^2}}} Factor the numerator.



{{{((n+1)(n-1))/((n+1)(n+1))}}} Expand the denominator.



{{{(highlight((n+1))(n-1))/(highlight((n+1))(n+1))}}} Highlight the common terms.



{{{(cross((n+1))(n-1))/(cross((n+1))(n+1))}}} Cancel out the common terms.



{{{(n-1)/(n+1)}}} Simplify



So {{{(n^2-1)/(n+1)^2}}} simplifies to {{{(n-1)/(n+1)}}} 



In other words, {{{(n^2-1)/(n+1)^2=(n-1)/(n+1)}}} where {{{n<>-1}}}