Question 123693
{{{((p^2 - q^2)/(p+q))/((p-q)/(p+q))}}}



Just like dividing any other fractions, you invert and multiply:


{{{((p^2 - q^2)/(p+q))*((p+q)/(p-q))}}}



{{{(p+q)/(p+q)=1}}}, so:



{{{((p^2 - q^2)/cross((p+q)))*(cross((p+q))/(p-q))}}}
{{{((p^2 - q^2)/(p-q))}}}


Now, remember the factorization of the difference of two squares:  {{{a^2-b^2=(a+b)(a-b)}}}


{{{((p^2 - q^2)/(p-q))}}}



{{{((p+q)(p-q))/(p-q)}}}


But {{{(p-q)/(p-q)=1}}}, so 


{{{((p+q)cross(p-q))/cross(p-q)}}}
{{{p+q}}}


Super-double-plus extra credit.  {{{((p^2 - q^2)/(p+q))/((p-q)/(p+q))=p+q}}}for all real numbers p and q with what restriction?