Need explanation for the solution of this limit with cubic root.

Question

I found this example but don't know how the author solved it, is it a conjugate multiplication. I tried that method but didn't work.

I don't know how it's solved in that way?

Could anyone explain the solution method for this limit example?

Answer 1

i would Substitute $$\sqrt[3]{y}=a$$ then you will get $$\frac{1-a^3}{1-a}=-\frac{\left( a-1 \right) \left( {a}^{2}+a+1 \right)}{1-a}$$ and this is $$a^2+a+1$$ and since $$y$$ tends to $1$ the variable $a$ must tends to one, and the searched Limit is $3$