Evaluating $\lim_{x\to2}\frac{\sqrt[3]{x^2-x-1}-\sqrt{x^2-3x+3}}{x^3-8}$ without Hopital rule I want to evaluate this limit without applying Hopital rule,$$\lim_{x\to2}\frac{\sqrt[3]{x^2-x-1}-\sqrt{x^2-3x+3}}{x^3-8}$$
After expanding the denominator I got,
$$\frac{1}{12}\lim_{x\to2}\frac{\sqrt[3]{x^2-x-1}-\sqrt{x^2-3x+3}}{x-2}$$Here  I can introduce numerator as $f(x)$ And say the limit is in the form $\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}=f'(2)$ But it is very similar to Hopital rule (if not the same).
I noticed that the fraction is in the form $2.\frac{\sqrt[3]A-\sqrt B}{A-B}$ or  $2.\frac{M^2-N^3}{M^6-N^6}$ but not sure if this helps.

use  $(a^6 - b^6) = (a-b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)$
Let $a = \sqrt [3]{x^2 - x - 1}, b = \sqrt{x^2 - 3x + 3}$
that is where we have $\frac {a-b}{x-2}$ we multiply top and bottom by $(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)$ giving
$\frac {a^6-b^6}{(x-2)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)}$
$\frac {(x^2 - x - 1)^2 - (x^2 - 3x+3)^3}{(x-2)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)}$
Multiply out the numerator and combine terms.  If you haven't made an aglebra error, what you have will be something that divides by $(x-2).$  Factor out $(x-2)$ and cancel out the $(x-2)$ in the denominator.
Plug $x=2$ to evaluate the numerator.
And as $x$ approaches $2,$ $a,b$ approach $1,$ making the denominator equal $6$
