Question 185954
whats the multiplicative inverse of  {{{(x^2+y^2)/(x-y)}}}

That asks the question:

What do you have to multiply {{{(x^2+y^2)/(x-y)}}} by to get 1?

Answer: You have to multiply it by the reciprocal of itself
in order to get 1.  So the answer is its reciprocal, namely,

{{{(x-y)/(x^2+y^2)}}}

To show why that's the answer, we multiply the original by it

{{{((x^2+y^2)/(x-y))((x-y)/(x^2+y^2))}}} 

Then we cancel:

{{{(cross((x^2+y^2))/(x-y))((x-y)/cross((x^2+y^2)))}}}

{{{(cross((x^2+y^2))/(cross(x-y)))((cross(x-y))/cross((x^2+y^2)))}}}

and that just gives {{{1}}}

Edwin</pre>