calculus - How to solve this limit without L'Hospital?

How can we solve this limit without using L'Hospital rule? I tried using some other methods but can't get the answer.

$$\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt[]{x}-1}$$

Answer

Let $x=t^6$. Note that

$$\lim_{x \to 1 }\frac{\sqrt[3]{x}-1}{\sqrt[]{x}-1}=\lim_{t \to 1} \frac{t^2-1}{t^3-1}=\lim_{t \to 1}\frac{t+1}{t^2+t+1}$$
So the limit is $\frac{2}{3}$.