How did my teacher gets this result? Algebra/Limits $$\begin{align*}&\lim_{x\to0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}=\\
&\lim_{x\to0}\left(\frac{(1+x)-(1-x)}{(1+x)-(1-x)}\cdot\frac{\sqrt[3]{(1+x)^2}+\sqrt[3]{(1+x)(1-x)}+\sqrt[3]{(1-x)^2}}{\sqrt{1+x}+\sqrt{1-x}}\right)
\end{align*}$$
My teacher was using this to calculate the value of the first limit as seen above. 
I am not sure how he pull this huge rabbit out of the hat XD.
Also I am not sure why is it useful...

To get from the first line to the second line, your teacher is multiplying by
$$
\frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \cdot \frac{\sqrt[3]{(1+x)^2} + \sqrt[3]{(1+x)(1-x)} + \sqrt[3]{(1-x)^2}}{\sqrt[3]{(1+x)^2} + \sqrt[3]{(1+x)(1-x)} + \sqrt[3]{(1-x)^2}}.
$$
Why did he make this choice?  He is using the identities
$$
(\sqrt{a} - \sqrt{b}) \cdot (\sqrt{a} + \sqrt{b}) = a - b
$$
and
$$
(\sqrt[3]{a} - \sqrt[3]{b}) \cdot (\sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}) = a - b.
$$
As with many limit problems, the reason for these manipulations is to try to obtain an algebraic cancellation to deal with the fact that both the numerator and denominator are approaching zero. 
