Question 138272
In general,
{{{(a+b)(c+d)=ac+ad+bc+bd}}}
Specifically in your case,
{{{(sqrt(2)+1)(sqrt(2)-1)=(sqrt(2))(sqrt(2))+sqrt(2)(-1)+(1)(sqrt(2))+(1)(-1)}}}
{{{(sqrt(2)+1)(sqrt(2)-1)=2-sqrt(2)+(sqrt(2))-1}}}
{{{(sqrt(2)+1)(sqrt(2)-1)=2-1}}}
{{{(sqrt(2)+1)(sqrt(2)-1)=1}}}