<s>[INST] 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 $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$?
 [/INST] 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
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