To solve this, remember that when multiplying variables, exponents are added.  When raising a power to a power, exponents are multiplied.  Thus:

$\displaystyle (x^{2})^{\frac{1}{3}}(x^{-2})^{\frac{1}{2}}(x^{4})^{\frac{1}{3}} =$

$\displaystyle x^{\frac{2}{3}}x^{-1}x^{\frac{4}{3}} = x^{\frac{2}{3}-1+\frac{4}{3}} = x$

Since $\displaystyle 18 = 2 \cdot 3 ^{2}$, we can multiply 18 by $\displaystyle 2^{2} \cdot 3 = 12$ to yield the lowest possible perfect cube:

$\displaystyle 18 \cdot 12 = 216 = 6 ^{3}$

Therefore, to rationalize the denominator, we multiply both nuerator and denominator by $\displaystyle \sqrt[3]{12}$ as follows:

$\displaystyle \frac{6}{\sqrt[3]{18} }$

$\displaystyle = \frac{6 \cdot \sqrt[3]{12} }{\sqrt[3]{18} \cdot \sqrt[3]{12} }$

$\displaystyle = \frac{6 \cdot \sqrt[3]{12} }{\sqrt[3]{18 \cdot 12 } }$

$\displaystyle = \frac{6 \sqrt[3]{12} }{\sqrt[3]{216 } }$

$\displaystyle = \frac{6 \sqrt[3]{12} }{6 }$

$\displaystyle = \sqrt[3]{12}$

Begin by getting a prime factor form of the contents of your root.

$\displaystyle 120000 = 12 * 10000 = 2^2*3*10^4$

Applying some exponent rules makes this even faster:

$\displaystyle 10^4 = (2*5)^4=2^45^4$

Put this back into your problem:

$\displaystyle 2^2*3*2^4*5^4=2^6*3*5^4$

Returning to your radical, this gives us:

$\displaystyle \sqrt[3]{2^6*3*5^4}$

Now, we can factor out $\displaystyle 2$ sets of $\displaystyle 2^3$ and $\displaystyle 1$ set of $\displaystyle 5^3$.  This gives us:

$\displaystyle 2^2*5\sqrt[3]{3*5}=20\sqrt[3]{15}$

Begin by factoring the contents of the radical:

$\displaystyle 15625 = 25*625 = 25*25*25=5*5*5*5*5*5=5^6$

This gives you:

$\displaystyle \sqrt[4]{5^6}$

You can take out $\displaystyle 1$ group of $\displaystyle 5^4$.  That gives you:

$\displaystyle 5\sqrt[4]{5^2}=5\sqrt[4]{25}$

Using fractional exponents, we can rewrite this:

$\displaystyle 5*5^\frac{2}{4}$

Thus, we can reduce it to:

$\displaystyle 5*5^\frac{1}{2}$

Or:

$\displaystyle 5\sqrt{5}$

To simplify $\displaystyle \sqrt{125}+\sqrt{50}$, find the common factors of both radicals.

$\displaystyle \sqrt{125}= \sqrt{25\cdot5}=\sqrt{25}\cdot \sqrt5=5\sqrt5$

$\displaystyle \sqrt{50}=\sqrt{25\cdot 2}=\sqrt{25}\cdot \sqrt2=5\sqrt2$

Sum the two radicals.

The answer is:  $\displaystyle 5\sqrt5+5\sqrt2$

To take the cube root of the term on the inside of the radical, it is best to start by factoring the inside:

$\displaystyle \sqrt[3]{x^3\cdot x^2\cdot y^6\cdot 27}$

Now, we can identify three terms on the inside that are cubes:

$\displaystyle x^3, y^6, 27$

We simply take the cube root of these terms and bring them outside of the radical, leaving what cannot be cubed on the inside of the radical.

$\displaystyle x\cdot y^2 \cdot 3 \cdot \sqrt[3]{x^2}$

Rewritten, this becomes

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

Simplify both radicals by rewriting each of them using common factors.

$\displaystyle \sqrt{32} = \sqrt4 \cdot \sqrt{4}\cdot \sqrt{2} = 4\sqrt2$

$\displaystyle \sqrt{8} = \sqrt4 \cdot \sqrt2 = 2\sqrt2$

Multiply the two radicals.

$\displaystyle 4\sqrt2\cdot 2\sqrt2 = 4\cdot 2 \cdot 2 = 16$

The answer is:  $\displaystyle 16$

In order to simplify this radical, rewrite the radical using common factors.

$\displaystyle -\sqrt{ 120} = -\sqrt{2} \cdot \sqrt{2}\cdot \sqrt{5} \cdot \sqrt{6}$

Simplify the square roots.  

$\displaystyle -2\cdot \sqrt{5} \cdot \sqrt{6}$

Multiply the terms inside the radical.

The answer is:  $\displaystyle -2\sqrt{30}$

Break down the two radicals by their factors.

$\displaystyle \sqrt{30} \cdot \sqrt{15} = (\sqrt{3}\times\sqrt{2}\times \sqrt{5})\cdot(\sqrt{3}\times \sqrt{5})$

A square root of a number that is multiplied by itself is equal to the number inside the radical.

$\displaystyle \sqrt{3}\cdot \sqrt{3}=3$

$\displaystyle \sqrt{5}\cdot \sqrt{5}=5$

Simplify the terms in the parentheses.

$\displaystyle (\sqrt{3}\times\sqrt{2}\times \sqrt{5})\cdot(\sqrt{3}\times \sqrt{5}) = 3\times 5 \times \sqrt2$

The answer is:  $\displaystyle 15\sqrt2$

The first term is already simplified.  The second and third term will need to be simplified.

Write the common factors of the second radical and simplify.

$\displaystyle 6\sqrt{50} =6\sqrt{25\times 2} = 6\sqrt{25}\times \sqrt{2} = 6(5)\sqrt2 = 30\sqrt2$

Repeat the process for the third term.

$\displaystyle 9\sqrt{500} = 9\sqrt{100\times 5} = 9\sqrt{100}\times \sqrt{5} = 9(10)\sqrt{5} = 90\sqrt5$

Rewrite the expression.

$\displaystyle 3\sqrt{5}+6\sqrt{50}+ 9\sqrt{500}= 3\sqrt{5}+30\sqrt2+90\sqrt5$

Combine like-terms.

The answer is:  $\displaystyle 30\sqrt2+93\sqrt5$