Question 1090261
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factor (1)/(x^(2))-(1)/(x)=6
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<pre>
Introduce new variable  y = {{{1/x}}}.      (1)


Then your equation will take the form

{{{y^2 - y}}} = 6,    or, equivalently,

{{{y^2 - y - 6}}} = 0.          (2)

The left side quadratic polynomial can be easily factored 

(y-3)*(y+2) = 0.      (3)


It gives the roots of the equation  y = 3  and  y = -2.


Let us consider both cases.


1.  If y = 3,   then  x = {{{1/y}}} = {{{1/3}}},  due to (1).


2.  If y = -2,  then x = {{{1/y}}} = {{{-1/2}}}.


<U>Answer</U>.  The original equation has the roots  {{{1/3}}}  and  {{{-1/2}}}.
</pre>

Solved.



<U>Notice</U>. &nbsp;The formulation of the problem in your post is not exactly adequate.


The ideally balanced formulation is &nbsp;<U>THIS</U>:


<pre>
     Solve the equation  (1)/(x^(2))-(1)/(x)=6  using factoring.
</pre>