$\displaystyle \frac{3}{1-\sqrt{2}}$ To get rid of the radical in the denominator, we need to multiply top and bottom by the conjugate which is the opposite sign of the radical expression. This would be $\displaystyle 1+\sqrt{2}$.

$\displaystyle \frac{3}{1-\sqrt{2}}\cdot\frac{1+\sqrt{2}}{1+\sqrt{2}}=\frac{3+3\sqrt{2}}{1-2}=\frac{3+3\sqrt{2}}{-1}$ Remember to use FOIL when multiplying out the denominators. Now, with out answer, we can distribute out the $\displaystyle -1$. 

$\displaystyle \frac{3+3\sqrt{2}}{-1}=-3-3\sqrt{2}$

$\displaystyle 6\sqrt{3}*2\sqrt{8}$

Multiply the outer numbers first:

$\displaystyle 12(\sqrt{3}*\sqrt{8})$

Combine radicals:

$\displaystyle 12\sqrt{24}$

Simplify the radical:

$\displaystyle 12\sqrt{6*4}$

$\displaystyle \mathbf{24\sqrt{6}}$

When multiplying radicals, we can just multiply the values inside the radicand.

$\displaystyle \sqrt{3}*\sqrt{7}=\sqrt{3*7}=\sqrt{21}$

When multiplying radicals, we can just multiply the values inside the radicand.

$\displaystyle \sqrt{3}*\sqrt{27}=\sqrt{3*27}=\sqrt{81}=9$

When multiplying radicals, we can just multiply the values inside the radicand.

$\displaystyle \sqrt{5}*\sqrt{15}=\sqrt{5*15}=\sqrt{75}$ We can simplify this by finding a perfect square.

$\displaystyle \sqrt{75}=\sqrt{25}*\sqrt{3}=5\sqrt{3}$

When dealing with radicals in the denominator, we simplify it by multiplying top and bottom by the radical.

$\displaystyle \frac{4}{\sqrt{2}}=\frac{4}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}$

We can simplify this fraction. We can combine the radicals into a giant radical expression.

$\displaystyle \frac{\sqrt{6}}{\sqrt{12}}=\sqrt{\frac{6}{12}}=\sqrt{\frac{1}{2}}$

When dealing with radicals in the denominator, we simplify it by multiplying top and bottom by the radical.

$\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

When multiplying radicals, we can just multiply the values inside the radicand.

$\displaystyle \sqrt[3]{4}*\sqrt[3]{16}=\sqrt[3]{64}$ This can be simplified to $\displaystyle 4$ which is the cubic root of the answer.

We can simplify $\displaystyle \sqrt{50}$ by finding a perfect square.

$\displaystyle \sqrt{50}=\sqrt{25}*\sqrt{2}=5\sqrt{2}$ Next we can reduce to $\displaystyle \frac{2}{\sqrt{2}}$.

When dealing with radicals in the denominator, we simplify it by multiplying top and bottom by the radical.

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

The first step I'd recommend is to multiply everything and put it all underneath one radical: $\displaystyle \sqrt{16x^3w^4}$. Then, recall that for every two of the same term, cross them out underneath the radical and put one of them outside. Attack each term separately: $\displaystyle \sqrt{16}=4$, $\displaystyle \sqrt{x^3}=x\sqrt{x}$, and $\displaystyle \sqrt{w^4}=w^2$. Put those all together to get: $\displaystyle 4xw^2\sqrt{x}$.