Limit $\lim_{x \to 2^{-}} \left ( \frac{1}{\sqrt[3]{x^{2} -3x+2}} + \frac{1}{\sqrt[3]{x^{2} -5x+6}} \right )$ $$\lim_{x \to 2^{-}} \left (  \dfrac{1}{\sqrt[3]{x^{2} -3x+2}} + \dfrac{1}{\sqrt[3]{x^{2} -5x+6}} \right )$$
I've tried using the $A^3-B^3$ identity, but that doesn't help. Also, I tried multiplying every fraction with $\sqrt[3]{A^2}$ to get rid of the roots in the denominator, but that doesn't help either. Can someone suggest a solution? Thanks.

Note that
\begin{align}
x^2-3x+2&=(2-x)(1-x)=(2-x)(2-x-1)
\\[4px]
x^2-5x+6&=(2-x)(3-x)=(2-x)(2-x+1)
\end{align}
Set $2-x=t^3$, so your expression becomes
$$
\frac{1}{\sqrt[3]{t^3(t^3-1)}}+\frac{1}{\sqrt[3]{t^3(t^3+1)}}
=\frac{1}{t}\frac{\sqrt[3]{t^3+1}+\sqrt[3]{t^3-1}}{\sqrt[3]{t^6-1}}
$$
and the limit is
$$
\lim_{t\to0^+}\frac{1}{\sqrt[3]{t^6-1}}\frac{\sqrt[3]{t^3+1}+\sqrt[3]{t^3-1}}{t}
$$
The first fraction has limit $-1$, whereas the second fraction has limit $f'(0)$, where $f(t)=\sqrt[3]{t^3+1}+\sqrt[3]{t^3-1}$. Since
$$
f'(t)=3t^2(t^3+1)^{-2/3}+3t^2(t^3-1)^{-2/3}
$$
we have $f'(0)=0$.
Alternatively,
\begin{align}
\frac{1}{\sqrt[3]{t^3(t^3-1)}}+\frac{1}{\sqrt[3]{t^3(t^3+1)}}
&=\frac{(1+t^3)^{-1/3}-(1-t^3)^{-1/3}}{t}\\[6px]
&=\frac{1-\frac{1}{3}t^3-1-\frac{1}{3}t^3+o(t^3)}{t}\\[6px]
&=-\frac{2}{3}t^2+o(t^2)
\end{align}
