If $$x = (\surd5) + 1$$ and $$y = (\surd5) - 1$$, then what is the value of $$\left(\frac{x^2}{y^2}\right) + \left(\frac{y^2}{x^2}\right) + 4[\left(\frac{x}{y}\right) + \left(\frac{y}{x}\right)] + 6$$?

question can be rewritten in the form of
$$\frac{(x^{2}+y^{2})^{2}-2x^{2}y^{2}}{(xy)^{2}}+4(\frac{(x+y)^{2}-2x^{2}y^{2}}{x^{2}y^{2}})+6$$
xy=4
x+y=$$2\sqrt{5}$$
$$(x^{2}+y^{2})^{2}-2x^{2}y^{2}$$=$$12^{2}-2(16)$$
=144-32
=112
$$(xy)^{2}$$=$$4^{2}$$=16
($$\frac{(x+y)^{2}-2x^{2}y^{2}}{x^{2}y^{2}}$$)=12/4
=3
$$\frac{(x^{2}+y^{2})^{2}-2x^{2}y^{2}}{(xy)^{2}}+4(\frac{(x+y)^{2}-2x^{2}y^{2}}{x^{2}y^{2}})+6$$=(112/16)+4(3)+6
=7+12+6
=25