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...

You should use the following identities:
$$a^2-b^2=(a-b)(a+b)$$
$$c^3-d^3=(c-d)(c^2+cd+d^2)$$
(prove by opening brackets) for 
$$a=\sqrt{1+x}, \hspace{5pt} b=\sqrt{1-x}, \hspace{5pt} c=\sqrt[3]{1+x}, \hspace{5pt} d=\sqrt[3]{1-x}$$
Then you multiply your function in the limit by $\frac{a+b}{c^2+cd+d^2}\cdot \frac{c^2+cd+d^2}{a+b}$, which leaves an equal expression. This is useful since instead of getting $\frac00$, you get an expression where you can substitute $x=0$.
