Question 443201
Notice that 4x^4 and 9y^4 are perfect squares.

(2x^2)^2 = 4x^4
(3y^2)^2 = 9y^4

We have the difference of squares:

Then,  {{{(4x^4 - 9y^4) = (2x^2 + 3y^2)(2x^2-3y^2)}}}

Noticing the same fact that 4x^4 and 9y^4 are perfect squares, and that (2*3)*2  = 12, we can say that {{{(4x^4 + 12x^2y^2 +9y^4) = (2x^2+3y^2)^2}}}

So we have {{{((2x^2 + 3y^2)(2x^2-3y^2))/(2x^2+3y^2)^2}}}

{{{(cross((2x^2 + 3y^2))(2x^2-3y^2))/(2x^2+3y^2)^cross(2)}}}
{{{(2x^2-3y^2)/(2x^2+3y^2)}}}