How to factor $(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$ to get $\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$? 
How can I factor this: $$\left(1+\frac{1}{x}\right)\left(-\frac{6}{x^2}\right)+\left(\frac{6}{x^3}\right)\left(1+\frac{1}{x}\right)^2$$
in order to get this result: $$\frac{6}{x^3}\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)$$

I have tried to factor out the $(1+\frac{1}{x})$ use the distributive property, and failed multiple times. Every online calculator I tried gives me different results.

Eqn 1
$$(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$$
$$(-\frac{6}{x^2})+(-\frac{6}{x^3})+(\frac{6}{x^3})(1+\frac{1}{x^2}+\frac{2}{x})$$
$$(-\frac{6}{x^2})+ (-\frac{6}{x^3}) + \frac{6}{x^3}+\frac{6}{x^5}+\frac{12}{x^4}$$
$$\frac6{x^2}[\frac{1}{x^3}+\frac{2}{x^2}-1]$$
Eqn 2
$$\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$$
$$(\frac{6}{x^3}+\frac{6}{x^4})(1+\frac2{x})$$
$$\frac6{x^3}[1+\frac3{x} +\frac2{x^2}]$$
On comparision we know that both the equations are different and there is is no way to convert Eqn 1 to Eqn 2.
