Use the multiplication property of radicals to split the fourth roots as follows:

$\displaystyle \rightarrow \frac{3\sqrt[4]{16}\sqrt[4]{2}}{2\sqrt[4]{81}\sqrt[4]{2}}$

Simplify the new roots:

$\displaystyle \rightarrow \frac{3(2)\sqrt[4]{2}}{2(3)\sqrt[4]{2}}$

$\displaystyle \rightarrow \frac{6\sqrt[4]{2}}{6\sqrt[4]{2}}$

$\displaystyle \rightarrow 1$

Use the multiplication property of radicals to split the perfect squares as follows:

$\displaystyle \sqrt{100}\sqrt{3}\sqrt{x^{2}}\sqrt{x}$

Simplify roots,

$\displaystyle 10\sqrt{3}\ast x\sqrt{x} = 10x\sqrt{3x}$

To simplify radicals, we need to factor the expression inside the radical. A radical can only be simplified if one of the factors has a square root that is an integer.

For this problem, we'll first find all of the possible radicals of 12: 1 & 12, 2 & 6, and 3 & 4. Then we look at each factor and determine if any of them has a square root that is an integer. The only one that does is 4, which has a square root of 2. We can rewrite the radical as $\displaystyle \sqrt{3\cdot 4}$ which can also be written as $\displaystyle \sqrt{3}\cdot \sqrt{4}$. Taking the squareroot of 4, we come to the answer: $\displaystyle 2\sqrt{3}$.

In order to simplify each radical, we must find the factors of its radicand that have a whole number as a square root, which will allow us to take the square root of that factor out of the radical. We start by factoring each radicand, looking for any factors that have a neat whole number as a square root:

$\displaystyle 4\sqrt{75}-2\sqrt{147}+\sqrt{12}=4\sqrt{3\cdot25}-2\sqrt{3\cdot49}+\sqrt{3\cdot4}$

After factoring each radicand, we can see that there is a perfect square in each: 25 in the first, 49 in the second, and 4 in the third. Because these factors are perfect squares, we can easily take their square root out of the radical, which then gets multiplied by the coefficient already in front of the radical:

$\displaystyle 4(5\sqrt{3})-2(7\sqrt{3})+(2\sqrt{3})=20\sqrt{3}-14\sqrt{3}+2\sqrt{3}$

After simplifying each radical, we're left with the same value of $\displaystyle \sqrt{3}$ in each term, so we can now add all of our like terms together to completely simplify the expression:

$\displaystyle 20\sqrt{3}-14\sqrt{3}+2\sqrt{3}=8\sqrt{3}$

In order to solve this equation, we must see how many perfect cubes we can simplify in each radical.

$\displaystyle \sqrt[3]{64x^5y^8z^6}$

First, let's simplify the coefficient under the radical. $\displaystyle 64$ is the perfect cube of $\displaystyle 4$. Therefore, we can remove $\displaystyle 64$ from under the radical, and what we have instead is:

$\displaystyle 4\sqrt[3]{x^5y^8z^6}$

Now, in order to remove variables from underneath the square root symbol, we need to remove the variables by the cube. Since radicals have the property

$\displaystyle \sqrt{x^2}=\sqrt{x}\cdot\sqrt{x}$

we can see that

$\displaystyle 4\sqrt[3]{x^5y^8z^6}=4\sqrt[3]{x^3x^2y^6y^2z^6}$

With the expression in this form, it is much easier to see that we can remove one cube from $\displaystyle x$, two cubes from $\displaystyle y$, and two cubes from $\displaystyle z$, and therefore our solution is:

$\displaystyle 4xy^2z^2\sqrt[3]{x^2y^2}$

\(\displaystyle Factor\; 396\; under\; the\; radical\; sign\)\(\displaystyle \sqrt{396}\)  =  \(\displaystyle \sqrt{36\times 11}\)  =  \(\displaystyle \sqrt{6^{2}\times 11}\)

=\(\displaystyle \sqrt{6^2}\sqrt{11}\)
=\(\displaystyle 6\sqrt{11}\)

$\displaystyle \small \sqrt{128}$

Find the factors of 128 to simplify the term.

$\displaystyle \small 64*2=128$

We can rewrite the expression as the square roots of these factors.

$\displaystyle \sqrt{128}=\sqrt{64}*\sqrt{2}$

Simplify.

$\displaystyle \sqrt{64}*\sqrt{2}=8\sqrt{2}$

$\displaystyle \small \sqrt{50}$

Start by finding factors for the radical term.

$\displaystyle \small 25*2=50$

We can rewrite the radical using these factors.

$\displaystyle \small \sqrt{25}*\sqrt{2}=\sqrt{50}$

Simplify the first term.

$\displaystyle 5*\sqrt{2}=5\sqrt{2}$

Always work the math under the radical before simplifying. We can't do any math so let's see if it's factorable. This isn't factorable either so the answer is just the problem stated. 

$\displaystyle \sqrt{x^2+y^2}\neq x+y$ If you don't believe it, let $\displaystyle x= 2$ and $\displaystyle y = 3$

$\displaystyle \sqrt{2^2+3^2}=\sqrt{13}\neq2+3$

Let's analyze each statement.

I.$\displaystyle \sqrt{a^2+b^2}=a+b$

Let's try to factor. This isn't factorable so this statement is usually false, NOT ALWAYS true.

$\displaystyle \sqrt{a^2+b^2}\neq a+b$ If you don't believe it, let $\displaystyle x= 3$ and $\displaystyle y = 4$

$\displaystyle \sqrt{3^2+4^2}=\sqrt{25}=5\neq3+4$

The only time this is true is if $\displaystyle a$ or $\displaystyle b$ were $\displaystyle 0$ and the other variable was a perfect square. 

II. $\displaystyle \sqrt{a^2}=a$

Let's say $\displaystyle a=2$. $\displaystyle \sqrt{2^2}=2$ This is very true HOWEVER, what if $\displaystyle a = -2$. $\displaystyle \sqrt{(-2)^2}\neq-2$. Square roots don't generate negative values. Remember to do the math inside the radicand before simplifying. Only positive values and zero are possible and since there is no restriction on $\displaystyle a$, all assumptions are based on $\displaystyle a$ being any real number. So we can elminate this statement since question is asking ALWAYS true. 

III. The smallest integer in a radicand that generates a plausible, real number and smallest value is 0. 

From the second statement reasoning, "only positive values and zero are possible", this confirms that this statement is always true. Integers are whole numbers found on a number line. Real numbers are numbers found on a number line including all rational numbers (integers that can easily be fractions) and irrational numbers(values that can't be written as fractions). Remember, a negative number in a square root creates imaginary numbers (numbers including $\displaystyle i$). Even if you decide to say $\displaystyle \sqrt{(-1)^2}=1$, it doesn't make statement III false. $\displaystyle 1$ is greater than $\displaystyle 0$ even though $\displaystyle -1$ is a smaller integer than $\displaystyle 0.$

Therefore III only is the correct answer.