Trying to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$ to be $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ I am asked to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$. The solution is provided as $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Here is my working:
$\frac{\sqrt{8}}{1-\sqrt{3x}}$ = $\frac{\sqrt{8}}{1-\sqrt{3x}}$ * $\frac{1+\sqrt{3x}}{1+\sqrt{3x}}$ = $\frac{1+\sqrt{8}\sqrt{3x}}{1-3x}$ = $\frac{1+\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{3x}}{1-3x}$ = $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Is $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$?
 A: $$\frac{\sqrt{8}}{1-\sqrt{3x}} = \frac{\sqrt{8}}{1-\sqrt{3x}} \cdot \frac{1+\sqrt{3x}}{1+\sqrt{3x}} = \color{blue}{\frac{\sqrt{8}\cdot\left({1+\sqrt{3x}}\right)}{1-3x}} = \frac{\sqrt{8}+\sqrt{24x}}{1-3x}$$
$$= \frac{\sqrt{2^3}+\sqrt{2^3\cdot3x}}{1-3x} = \frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$$
Notice the step I highlighted in blue, which is where you made an error. You have to multiply $\sqrt{8}$ to $\left(1+\sqrt{3x}\right)$ completely. You multiplied it by only $\sqrt{3x}$ and added $1$, which wasn’t correct.
A: Observe that
$$
\sqrt{8}\times (1+\sqrt{3x})=\sqrt{8}+\sqrt{8}\times \sqrt{3x}
$$ and
$$
\sqrt{8}\times \sqrt{3}=\sqrt{24}=\sqrt{4\times 6}=2\sqrt{6}.
$$
A: Well, in rationalizing the denominator, we arrive at the intermediate step $\frac{\sqrt{8}+\sqrt{24x}}{1-3x}$ which simplifies to the provided solution. 
Your error comes in multiplying by the conjugate of the denominator. $\sqrt{8}*(1+\sqrt{3x})$ becomes $\sqrt{8}+\sqrt{24x}$. 
A: Write $$\frac{\sqrt{8}(1+\sqrt{3x})}{(1-\sqrt{3x})(1+\sqrt{3x})}$$ and this is
$$\frac{2\sqrt{2}(1+\sqrt{3x})}{1-3x}$$
A: Hints:$\sqrt8=\sqrt{4×2}=2\sqrt{2}.$ 
$\sqrt24=2\sqrt6$. (Why?)
$(1+\sqrt{3x})×(1-\sqrt{3x})=1-3x $.
A: The second equality is where you mess up - note that
$$\sqrt{8}\cdot(1+\sqrt{3x})=\sqrt{8}+\sqrt{8}\cdot\sqrt{3x}$$
by the distributive property of real numbers
