How to find $\sqrt{1+{4\over x}+{4\over x^2} }$? If   $$abx^2 = (a-b)^2(x+1)$$ then what is $$\sqrt{1+{4\over x}+{4\over x^2} }$$ 

(A) $a+b \over a-b$  (B)$a-b\over a+b$ (C) $a^2+ab$ (d) None

EDIT: What I've done is this:
$$abx^2=(a^2+b^2-2ab)(x+1)$$
=> $$abx^2 - (a^2+b^2+2ab)x - (a-b)^2=0$$
=> $$abx^2 - (a-b)^2x - (a-b)^2=0$$
but this is getting me nowhere.

$\textbf{Shortest Solution}$:
$$\frac{ab}{(a-b)^2}=\frac{(x+1)}{x^2}$$
$$\frac{ab}{(a-b)^2}=\frac{1}{x^2}+\frac{1}{x}$$
Multiplying $4$ both sides,
$$\frac{4ab}{(a-b)^2}=\frac{4}{x^2}+\frac{4}{x}$$
Adding $1$ both sides,
$$1+\frac{4ab}{(a-b)^2}=1+\frac{4}{x^2}+\frac{4}{x}$$
$$\frac{(a-b)^2+4ab}{(a-b)^2}=1+\frac{4}{x^2}+\frac{4}{x}$$
$$\frac{(a+b)^2}{(a-b)^2}=1+\frac{4}{x^2}+\frac{4}{x}$$
Taking Square root both sides,
$$\sqrt{\frac{(a+b)^2}{(a-b)^2}}=\sqrt{1+\frac{4}{x^2}+\frac{4}{x}}$$
$$\frac{(a+b)}{(a-b)}=\sqrt{1+\frac{4}{x^2}+\frac{4}{x}}$$
