Calculate the lim $\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}$ $$
\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}
$$
I multiply it with:
$$
\frac{ \sqrt{x} + 1}{\sqrt{x} + 1}
$$
And I get :
$$
\lim_{x\to 1} \frac{ \sqrt{x}^2 - 1^2}{(\sqrt[3]{x} - 1) * (\sqrt{x} + 1)}
$$
But the solution is still division by 0 and not possible, so I think I've made a mistake somewhere..

You basically want to insert $1$ written as a “conjugate” that removes the radical divided by itself. For $\sqrt{x}-1$ it is $\sqrt{x}+1$, for $\sqrt[3]{x}-1$ it is $\sqrt[3]{x^2}+\sqrt[3]{x}+1$. So you get
\begin{align}
\lim_{x\to1}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}&=
\lim_{x\to1}\left((\sqrt{x}-1)\frac{\sqrt{x}+1}{\sqrt{x}+1}\right)
  \left(\frac{1}{\sqrt[3]{x}-1}
    \frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\right)
\\[2ex]
&=\frac{x-1}{\sqrt{x}+1}\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{x-1}
\end{align}
and the indetermination goes away.
It's the same as Bill Dubuque's answer, actually, but without substitution.
