Evaluating $ \lim_{x\to \infty}x^p((x+1)^{1/3}+ (x-1)^{1/3} - 2x^{1/3})$ 
If $L = \lim_{x\to \infty}x^p((x+1)^{1/3}+ (x-1)^{1/3} - 2x^{1/3})$,
  where L is some non zero number, then $\dfrac{p^2}{L}= ? $

Attempt: 
I have tried binomial theorem for fractional indices but that does' not really help.  It's not possible to factorise or rationalise, so how do I go about solving it? 
 A: HINT
We have that by binomial series
$$(x+1)^{1/3}+ (x-1)^{1/3} - 2x^{1/3}=x^\frac13\left((1+1/x)^{1/3}+ (1-1/x)^{1/3} - 2\right)=$$
$$=x^\frac13\left(1+\frac1{3x}-\frac1{9x^2}+ 1-\frac1{3x}-\frac1{9x^2} - 2+o(1/x^2)\right)$$
A: If $n\in\mathbb {N} $ then we can show that $$\lim_{x\to 0}\frac{(1+x)^{1/n}-1-(x/n)}{x^2}=\frac{1-n}{2n^2}\tag{1}$$ using algebra alone. Just put $A=(1+x) ^{1/n},B=1+(x/n)$ and then $A-B=(A^n-B^n) /C$ where $A, B$ tend to $1$ and $C\to n$ and $$A^n-B^n=1+x-1-x-\frac{n(n-1)}{2n^2}x^2+o(x^2)$$ and thus it follows that $$\frac{A-B} {x^2}\to \frac{1-n}{2n^2}$$ as $x\to 0$ as desired.
The given limit condition can be transformed into $$L=\lim_{t\to 0^{+}}t^{5/3-p}\cdot\left(\frac{(1+t)^{1/3}-1-t/3}{t^2}+\frac{(1-t)^{1/3}-1+t/3}{t^2}\right) $$ Both the fractions above tend to $-1/9$ via formula $(1)$ and hence it follows that $$\lim_{t\to 0^{+}}t^{5/3-p}=-\frac{9L}{2}$$ Since $L\neq 0$ it is now obvious that $p=5/3,L=-2/9,p^2/L=-25/2$.
The above approach avoids the use of binomial theorem for rational index and instead uses the simpler version of binomial theorem for positive integral index. 
