This problem rewards those who see that roots and exponents are the same operations (roots are "fractional exponents"), and who therefore choose the easier order in which to perform the calculation. The trap here is to have you try to square 27. Not only is that labor-intensive, but once you get to 729 you then have to figure out how to take the cube root of that!

Because you can handle the root and the exponent in either order (were you to express this as a fractional exponent, it would be , which proves that the root and exponent are the same operations), you can take the cube root of 27 first if you want to, which you should know is 3. So at that point, your problem is what is ?" And you, of course, know the answer: it's 9.

It is important to be able to convert between root notation and exponent notation. The third root of a number (for example,  is the same thing as taking that number to the one-third power .

So when you see that you're taking the third root of , you can read that as  to the  power: 

This then allows you to apply the rule that when you take one exponent to another power, you multiply the powers:

This then means that you can express this as:

With roots, it is important that you are comfortable with factoring and with expressing roots as fractional exponents. A square root, for example, can be expressed as taking that base to the  power. Using that rule, the given expression, , could be expressed using fractional exponents as:

This would allow you to then add the exponents and arrive at:

Since that 2 in the denominator of the exponent translates to "square root," you would have the square root of : 

If you were, instead, to work backward from the answer choices, you would see that answer choice  factors to the given expression. If you start with:

You can express that as:

That in turn will factor to:

The first root then simplifies to , leaving you with: 

Therefore, as you can see, choice  factors directly back to the given expression.

Simplify each of the expressions to determine which satisfies the condition of the problem:

If you're looking for a number exactly  units to either side of , you need an absolute value equation that gives you answers of  and .

One way forward is to simply solve each equation for  using the two-case method, assuming first that the expression within the absolute value signs positive and then that it is negative and solving the resulting equation for each case.

If you do so, choice  becomes: , or  and , or , which is correct.

Choice  becomes: , or , which doesn't work. 

Choice  becomes: , or , which doesn't match. 

Choice  becomes: , or , which doesn't work.

You could also quickly solve this question if you remember the general way number line absolute value problems like this one work. Each one will be in the general form , where the "Distance" referenced is the distance from the center value to one of the two values for .

So in this case the "center value" would be  and the "distance" would be , yielding an equation of .

Any time you have an absolute value equation and a question that is asking for the smallest possible value for the expression, remember that absolute value expressions can be zero but can never be less than 0. There are two ways to solve a problem like the one given. The first is by finding the value of  such that the expression within the absolute value signs is zero and then finding the closest integer to that number. The second is by inspection (simply using brute force to find the value for xx that will minimize the value given). 

If you set , and that . Since , the closest integer value is , which will yield a value of .

The other way to approach this problem is by inspection, or through brute force. Try plugging in a few values for  to see what happens.

If  then the expression is 

If , then the expression is .

If x=2x=2 then the expression becomes .

Notice that the function decreases and then increases around , indicating that (for integer values, at least)  will yield the smallest result for .

This problem is a perfect candidate to test the answer choices, as the calculations required are not particularly difficult but the algebraic setup can be quite challenging to conceptualize. If you look for the two numbers that are 66 units away from −4−4, they are: 

Then plug those values in as  and see which equations work. For answer choice ,  does equal , and  also equals , so  satisfies the equation.

Even if you're unsure of where to start on this problem, you should have a head start. The problem is testing absolute values, and you should know that the result of any absolute value is always nonnegative, . So the answer choices that include an absolute value equalling a negative number must be incorrect: that just cannot be possible.

To test the remaining choices, consider that the numbers that are exactly two units away from -3 are -3+2 = -1, and -3-2 = -5.  When you plug these numbers in for  in the answer choices, only one is valid:

 gives you:

 --> 

 --> 

Therefore this absolute value satisfies the given situation and is correct.

It is important to recognize that absolute values must be nonnegative, . That means that for this given expression, the  can only go as low as , and then the second part of the expression asks you to add . So this expression can never equal zero: it's an absolute value added to 1, so the lowest this expression can be is 1.

To solve an absolute value like this, recognize that there are two outcomes inside an absolute value that would have it equal 3.  If the inside of an absolute value expression is 3, then the result is 3. Or if the inside of an absolute value expression equals -3, the absolute value will equal 3.  So you can solve this as two equations:

 and .

Solving for the first one, you have:

And solving for the second one, you have:

Therefore, the correct answer is: