This problem emphasizes an important concept about mixing exponents and roots: when you see exponents and roots mixed, be ready to treat roots as fractional exponents. That way all of your terms are in a common form, and you can directly apply exponent rules to all terms.

Here, that means that you'll treat  as . With that, the question then becomes:

Now you can use the exponent rule that  (when you divide exponents of the same base, you subtract the exponents) to turn this into:

While you might now look to immediately convert back to the form of the answer choices (which each use radical signs to represent roots),  is not an option. Which brings up another important point about mixing roots with exponents: generally when you're in exponent form you should stay there as long as possible, as the exponential form tends to provide you with more algebraic flexibility.

Here with  , you can pull out the  (or just ) to break apart the mixed-number exponent. You can rephrase this as , which then allows you to convert to an expression that looks like the answer choices. With the only fractional exponent left, , translating to , you can re-express the entire expression as .

This problem creates confusion for people because they are not comfortable rationalizing the cube root in the denominator. To see how to do this properly, first view the given expression like this:

and then ask what you would need to multiply the denominator by (and thus the numerator so that you're effectively multiplying by ) to make the expression under the root sign a perfect cube in the denominator. In this case, you must multiply top and bottom by  as shown below:

.

To simplify this multi-denominator expression, you should first multiply by one in the form  to eliminate the two denominators in the bottom of the expression. This multiplication results in the following:

Now you must multiply by one again to remove the two roots in the denominator. To do that you must recognize the difference of squares and multiply the top and the bottom by the conjugate of  as shown below:

Therefore, the correct answer is .

All of the exponent rules deal with multiplying, rather than adding, bases. In order to turn this into a multiplication question, we count apples (or chickens, or s, or whatever). How many ’s are there here? Three. This expression can be rewritten as . Now the exponent rules will apply; . The answer is .

To solve this problem we must recognize that  can be broken down into 

After breaking  into  we see that the  terms in numerator and denominator cancel out.

This leaves us with 

This problem tests your fluency with exponent rules, and gives you a helpful clue to guide you through using them. Here you may see that both 27 and 9 are powers of 3.   and . This allows you to express  as  and  as . Then you can simplify those exponents to get . Since when you divide exponents of the same base you subtract the exponents, you now have , and since  you really have: 

This then tells you that .

Note that had you not immediately seen to express all the numbers in this problem as powers of 3, the fact that the question asks for such a combination of variables, , should be your clue; you're given an exponent problem and asked for a subtraction answer, so that should get you thinking about dividing exponents of the same base to subtract the exponents, and at least give you some fodder for playing with exponent rules until you find a way to make progress.

This problem involves some creative factoring, and factoring is something you should always look to do whenever you see several exponents amidst some addition or subtraction. You cannot here factor any one thing out of all four terms, but you can group the terms to factor some common elements:

Can become:

And if you factor the common  term you have:

The only real number solution available is , so the correct answer is .

An important principle of exponents being tested here is that when you multiply/divide exponents of the same base, you add/subtract those exponents. Here you can do the corollary; if you had , you would add together those exponents to get . But in this case you're given the combined exponent  and may want to convert it to  so that you can factor:

 allows you to factor the  terms to get:

You can do the arithmetic to simplify , allowing you to then divide both sides by 3 and have:

So .

This problem hinges on your ability to recognize 16, 4, and 64 all as powers of 4 (or of 2).  If you make that recognition, you can use exponent rules to express the terms as powers of 4:

Since taking one exponent to another means that you multiply the exponents, you can simplify the numerator and have:

And then because when you divide exponents of the same base you can subtract the exponents, you can express this as:

This means that .

With this exponent problem, the key to getting the given expression in actionable form is to find common bases. Since both 9 and 27 are powers of 3, you can rewrite the given expression as:

When you've done that, you're ready to apply core exponent rules. When you take one exponent to another, you multiply the exponents. So the numerator becomes  and the denominator becomes . Your new fraction is:

Next deal with the negative exponents, which means that you'll flip each term over the fraction bar and make the exponent positive. This then makes your fraction:

Since when you divide exponents of the same base you subtract their exponents, this simplifies to .