We will need to rewrite the negative exponent as a fraction.

$\displaystyle -16^{-\frac{3}{4}} = -\frac{1}{16^{\frac{3}{4}}}$

Recall that the denominator of the power represents the root index, and the numerator of the power represents what the quantity is raised to.

$\displaystyle 16^{\frac{3}{4}} =( \sqrt[4]{16})^3 = (2)^3 = 8$

Replace the denominator.

$\displaystyle -\frac{1}{16^{\frac{3}{4}}} = -\frac{1}{8}$

The answer is:  $\displaystyle -\frac{1}{8}$

The denominator of the fractional exponent represents the index of the radical.

Rewrite the expression in radical form.  A radical with an index of 2 is simply the square root of a number.

$\displaystyle 27^\frac{1}{3}-16^{\frac{1}{2}} = \sqrt[3]{27}-\sqrt{16} = 3-4 = -1$

The answer is:  $\displaystyle -1$

The denominator represents the index of the square root.  The numerator is the  power that the quantity is raised to.

Rewrite the expression as a radical.

$\displaystyle 125^\frac{7}{3} =( \sqrt[3]{125})^7 = 5^7$

The answer is:  $\displaystyle 5^7$

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.

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