Condensation of an expression There is an expression: $(h+2p)^2(x+2y)^2 - 3p^2(x+2y)^2-3y^2(h+2p)^2+9p^2y^2$
Is there a way to simplify this to the form $l^2 + 4lk + k^2$? I tried to substitute $l = (h+2p)(x+2y)$ and $k= 3py$ and this takes care of the expressions at the two ends, but the expressions in the middle are stumping me.
Alternatively, one may try expressing it in the form $(l+2k)^2 - 3lk$, but I am unable to do this as well. However, I suspect that there is some relationship as $-3lk$ bears resemblance to $- 3p^2(x+2y)^2-3y^2(h+2p)^2$ if you factor out the $-3$. Is it possible to use Fibonnacci Identity in this endeavor?
EDIT: The original expression is $(h^2 + 4ph + p^2)(x^2 + 4xy + y^2)$ 
I converted this to $$((h+2p)^2-3p^2)((x+2y)^2-3y^2)$$
EDIT: I tried an alternative approach. I multiplied out this expression:
$$(h^2 + 4ph + p^2)(x^2 + 4xy + y^2)$$
and got the following:
$$h^2x^2 + p^2x^2 + h^2y^2 + p^2y^2 + 4hpx^2 + 4xyh^2 + 4xyp^2 +4hpy^2 + 16hpxy$$
(I'm leaving out some steps) This can be factored into :
$$(hx + py)^2+4(px+hy)(hx+py)+(px+hy)^2 +12hpxy$$
This would have been perfect except for the pesky $12hpxy$.


Answer:
  $$(h x-p y)^2 + 4(h x-p y)(h y+p x+4 p y) +(h y+p x+4 p y)^2 = (h^2 + 4p h + p^2)(x^2 + 4x y + y^2) $$

How I found this result:
Rewrite
$$(h^2 + 4p h + p^2)(x^2 + 4x y + y^2) = ((h+2p)^2-3p^2)((x+2y)^2-3y^2) = (A^2-3B^2)(C^2-3D^2)$$
where $A=h+2p$, $B=p$, $C=x+2y$ and $D=y$. Now, we have
$$(A^2-3B^2)(C^2-3D^2) = (AC+3BD)^2 - 3(AD + BC)^2 = \lbrace[AC+3BD-2(AD+BC)]+2(AD+BC)\rbrace^2 - 3(AD + BC)^2 $$
Let $l=AC+3BD-2AD-2BC$ and $k=AD + BC$. Now, $(l+2k)^2-3k^2=l^2 +4lk + k^2 $, which is what we wanted. Finally, substitute $A$, $B$, $C$ and $D$.
