$\displaystyle \sqrt{(16)(8)+(32)(20)}$

First, break down the components of the square root:

$\displaystyle \sqrt{(2^{4})(2^{3})+(2^{5})(2^{2})\times 5}$

Combine like terms. Remember, when multiplying exponents, add them together:

$\displaystyle \sqrt{(2^{7})+(2^{7})\times5}$

Factor out the common factor of $\displaystyle 2^7$:

$\displaystyle \sqrt{(2^{7})(1+5)}$

$\displaystyle \sqrt{(2^7)\times6}$

Factor the $\displaystyle 6$:

$\displaystyle \sqrt{(2^7)\times2\times3}$

Combine the factored $\displaystyle 2$ with the $\displaystyle 2^7$:

$\displaystyle \sqrt{(2^{8})\times3}$

Now, you can pull $\displaystyle \sqrt{2^8}$ out from underneath the square root sign as $\displaystyle 2^4$:

$\displaystyle 2^4\sqrt{3}$

$\displaystyle \sqrt{(27)(45)(125)}$

First, break down the component parts of the square root:

$\displaystyle \sqrt{(3^{3})(5\times 3^{2})(5^{3})}$

Combine like terms in a way that will let you pull some of them out from underneath the square root symbol:

$\displaystyle \sqrt{(5^{4})(3^4)(3)}$

Pull out the terms with even exponents and simplify:

$\displaystyle (5^{2})(3^{2})\sqrt{3}=225\sqrt{3}$

The cube root of $\displaystyle x$ is $\displaystyle x$ raised to the power of one third, or $\displaystyle x^{\frac{1}{3}}$. This raised to the power of twelve is

$\displaystyle \left (x^{\frac{1}{3}} \right )^{12} = x^{\frac{1}{3} \cdot 12} = x^{4}$

$\displaystyle 10,000 ^{-\frac{1}{3}}$

$\displaystyle = \frac{1}{10,000^{\frac{1}{3}}}$

$\displaystyle = \frac{1}{\sqrt[3]{10,000 }}$

$\displaystyle = \frac{1}{\sqrt[3]{1,000 }\cdot \sqrt[3]{1 0 }}$

$\displaystyle = \frac{1}{10\sqrt[3]{1 0 }}$

$\displaystyle = \frac{ \sqrt[3]{10 0 }}{10\sqrt[3]{1 0 }\cdot \sqrt[3]{10 0 }}$

$\displaystyle = \frac{ \sqrt[3]{10 0 }}{10 \cdot \sqrt[3]{1, 00 0 }}$

$\displaystyle = \frac{ \sqrt[3]{10 0 }}{10 \cdot 10}$

$\displaystyle = \frac{ \sqrt[3]{10 0 }}{10 0}$

The fourth root of $\displaystyle x$ is $\displaystyle x^{ \frac{1}{4}}$; the eighth power of this is $\displaystyle \left (x^{ \frac{1}{4}} \right )^{8} = x^{\frac{1}{4}\cdot 8} = x^{2}$.

The eighth root of $\displaystyle x$ is $\displaystyle x^{ \frac{1}{8}}$; the fourth power of this is $\displaystyle \left (x^{ \frac{1}{8}} \right )^{4} = x^{\frac{1}{8}\cdot 4} = x^{ \frac{1}{2}} = \sqrt{x}$.

The product of these expressions is $\displaystyle x^{2}\sqrt{x}$.

The square of $\displaystyle x$ is $\displaystyle x^{2}$.  The eighth root of $\displaystyle x^{2}$ is $\displaystyle x^{2}$ raised to the power of $\displaystyle \frac {1}{8}$, or 

$\displaystyle \left (x^{2} \right )^{\frac{1}{8}} = x ^{2 \cdot \frac{1}{8}} = x^{\frac{1}{4}}$

This is equivalent to $\displaystyle \sqrt[4]{x}$, the fourth root of $\displaystyle x$.

Each of the choices is a power of 10, so rewrite each choice as such:

$\displaystyle 10,000 = 10^{4}$

$\displaystyle 100,000 = 10^{5}$

$\displaystyle 1,000,000 = 10^{6}$

$\displaystyle 10,000,000 = 10^{7}$

$\displaystyle 100,000,000 = 10^{8}$

Since 10 itself does not have a rational square root, a necessary and sufficient condition for $\displaystyle 10^{N}$ to have a rational square root - that is, for

$\displaystyle \sqrt{10^{N} }= \left (10^{N} \right )^{\frac{1}{2}} = 10^{\frac{N}{2} }$ 

to be rational - is for $\displaystyle \frac{N}{2}$ to be an integer. This allows us to eliminate $\displaystyle 100,000 = 10^{5}$ and $\displaystyle 10,000,000 = 10^{7}$. 

Similarly, for $\displaystyle 10^{N}$ to have a rational cube root, $\displaystyle \frac{N}{3}$ must be an integer. This allows us to eliminate $\displaystyle 10,000 = 10^{4}$ and $\displaystyle 100,000,000 = 10^{8}$.

$\displaystyle 1,000,000 = 10^{6}$ is left. $\displaystyle 1,000,000$ is the correct choice.

The tenth power of the fifth root of $\displaystyle x$ can be found as follows:

The fifth root of $\displaystyle x$ is $\displaystyle \sqrt[5]{x} = x^{\frac{1}{5}}$; the tenth power of this is $\displaystyle \left ( x^{\frac{1}{5}} \right )^{10} = x^{\frac{1}{5}\cdot 10} = x^{2}$.

The tenth power of $\displaystyle x$ is $\displaystyle x^{10}$; the fifth root of this is $\displaystyle \left ( x ^{10} \right )^{\frac{1}{5}} = x^{10 \cdot \frac{1}{5}} = x^{2}$

The two expressions are equivalent, so they add up to $\displaystyle 2x^{2}$.

This question requires that you understand negative exponents and fractional exponents. To make a negative exponent positive, simply switch its location in the fraction, so this:

$\displaystyle x^{-\frac{3}{4}}$

becomes:

$\displaystyle \large \frac{1}{x^{\frac{3}{4}}}$

Then, we need to recognize how to rewrite fractional exponents. In a fractional exponent, the numerator stays as the power to which we are raising our base. The denominator becomes the index of our root. Thus, our fractional exponent becomes:

$\displaystyle \frac{1}{\sqrt[4]{x^3}}$

Four was the denominator of our fractional exponent, so it became the index of our root. In other words, something raised to the power of $\displaystyle \frac{1}{4}$ is the same thing as taking the fourth root of that something.

To solve this problem, we must remember exponent rules. The powers here need to be added, since the powers are both raising to the same number. The final answer is $\displaystyle 2^{n+n+1}$ or $\displaystyle 2^{2n+1}$.