Consider the equation x^3=x.  
How many values of x satisfy this equation? Many students will divide both sides by x to get 
x^2=1
Some will then take the square root of both sides of this equation to get
x=1
Looks like one solution.  However, it turns out that the original equation has 3 solutions.  In both steps we lost a solution.  To see this, let's resolve
x^3=x.
Subtract x from both sides:
x^3-x=0
Factor x from both terms:
x(x^2-1)=0
Factor again:
x(x-1)(x+1)=0
Now we use the fact that terms can only multiply to zero if some of them equal zero to get the three solutions:
x=0, x-1=0, and x+1=0, so that
x=0, x=1, and x=-1 are the three solutions.

So what went wrong?  Well it turns out that in the the first step at the very beginning of the post, when we divided by x, we could potentially be dividing by zero, which is not possible.  In doing so, we lost the x=0 solution by canceling out the factor of x.  

In the second step, where we took the square root of both sides, we also lost a solution because taking the square root only includes the positive square root.  There is a positive and a negative square root.

To sum up, when dividing by a variable, we could potentially lose the x=0 solution. To avoid this issue, there are a couple of approaches. One is to factor out the variable, like we did the second time around, rather than dividing it out and canceling. The other is to remember to write x=0 as a potential solution any time we divide by x.
Similarly for avoiding the square root issue, we could factor, or we could remember to write a +/- sign to include both the positive and negative roots.

One other note is that, going the other direction, 
Multiplying by 0 can potentially introduce false (aka extraneous) x=0 solutions, and squaring an equation can also similarly introduce a false solution.


For example, if we multiply both sides of
x=1 by x, we get
x^2=x, which now has 2 solutions, x=0 and x=1.  We introduced the extraneous x=0 solution by multiplying by x.
Similarly, if x = 1, and we square both sides, we get that x^2=1, which has 2 solutions: x=1 and x=-1.  We introduced the false extraneous negative solution by squaring.

Remember to plug the final solutions back in to the original equation to make sure no extraneous solutions were introduced.

The final summary:
Dividing or taking roots can drop solutions.
Multiplying or taking exponents can introduce extraneous solutions (so make sure to plug your solutions obtained to eliminate extraneous solutions).

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Many students often struggle with the question "What or how is the best way to choose variables?". One key to solving many math problems is in how the information is represented.  We want the variables to convey all the relevant information without being cumbersome to write repeatedly or to understand. The optimal choice of variables:

1. Is simple, easy to read, and has no unnecessary baggage,
2. Conveys the relevant information,
3. Doesn't require a key for you to remember what they represent.

For example, suppose we have two entrepreneurs, Alexis and Brandon, and we are to compare their revenues at two different times. How would you represent these quantities?

We could label the individuals A and B, and the time periods 1 and 2.  We could call the revenues R_A and R_B, but since we are not interested in any other quantities, we can drop the R and just call the revenues A and B, accomplishing 1).  Now since we are dealing with two different time periods, we can call Alexis's revenue in the first time period A1, Alexis's revenue in the second time period A2, and so on, so that we have A1, A2, B1, and B2.  Now we have variable representation that covers all relevant information, accomplishing 2), but that is not so complex that we need to refer to a key to remember what each variable represented, accomplishing 3).