Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$ 
Evaluate the limit
$$
\lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right)
$$

My Attempt:
To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B = \left(\sqrt[3]{x^3-x^2+1}\right)$. Now
$$
\begin{align}
2x^2 &= A^3-B^3\\
x &= \sqrt{\frac{A^3-B^3}{2}}
\end{align}
$$
So the limit becomes
$$\lim_{x\rightarrow \infty}\left(\sqrt{\frac{A^3-B^3}{2}}-A-B\right)$$
How can I complete the solution from this point?

Hint
Since $x$ goes to infinity, let us write $$\sqrt[3]{x^3+x^2+1}=x\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}}$$ and let us define $y=\frac{1}{x}+\frac{1}{x^2}$. Now, look at the Taylor expansion of $$\sqrt[3]{1+y}=1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)$$ and replace $y$ by its definition and expand to get finally $$\sqrt[3]{x^3+x^2+1}=x \Big(1+\frac{1}{3 x}-\frac{1}{9 x^2}+...\Big)\simeq x-\frac{1}{9 x}+\frac{1}{3}$$ Similarly, the same method should give $$\sqrt[3]{x^3-x^2+1} \simeq x-\frac{1}{9 x}-\frac{1}{3}$$
I am sure that you can take from here.
More intuitively, you could have noticed that, since $x$ goes to $\infty$,$$\sqrt[3]{x^3 \pm x^2+1}=x\sqrt[3]{1 \pm\frac{1}{x}+\frac{1}{x^2}}$$ behaves just as $x$ and then the limit of the expression is $0$. However, this approach allows you to show "how" the expression goes to $0$ (by positive or negative values).
