find a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}$ 
Let $\{x\} = x-\lfloor x\rfloor$ be the fractional part of $x$. Order the (real) solutions to $\sqrt{\lfloor x\rfloor \lfloor x^3\rfloor} + \sqrt{\{x\}\{x^3\}} = x^2$ with $x\ge 1$ from smallest to largest by $x_1,x_2,\cdots$. Provide a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}.$

First, I'm not even sure why there are infinitely many solutions to $\sqrt{\lfloor x\rfloor \lfloor x^3\rfloor} + \sqrt{\{x\}\{x^3\}} = x^2$. One inequality that might be useful is the Cauchy-Schwarz inequality. There's also the AM-GM inequality. We have $ac + bd \leq \sqrt{a^2+b^2}\sqrt{c^2+d^2}$ for all real numbers $a,b,c,d$ where $ac,bd\ge 0$ with equality iff $(a,b),(c,d)$ are proportional vectors in $\mathbb{R}^2$. Observe that $\lfloor x\rfloor$ always has the same sign as $x$, so $\lfloor x\rfloor \lfloor x^3\rfloor$ has the same sign as $x^4 \ge 0$. It might be useful to substitute $\{x\} = x-\lfloor x\rfloor$ into the original equation and simplify the result somehow. Also, it could be possible to write the given sum as a telescoping sum.

Let $u$ be a noncubic integer with $\lfloor\sqrt[3]{u}\rfloor = n \in \mathbb{Z}^+$. Assuming $\sqrt[3]{u} \le x < \sqrt[3]{u+1}$, we can reformulate the equation as follows.
$$ \sqrt{nu} + \sqrt{(x-n)(x^3-u)} = x^2 $$
Squaring both sides, we have
$$(\sqrt{n} x - \sqrt{u})^2 = 0$$
Hence the equation holds for $x = \sqrt{u/n}$ if $\sqrt[3]{u} \le \sqrt{u/n} < \sqrt[3]{u+1}$. Since the first inequality always holds, we only need to check which $u$ satisfies the second inequality. After calculating some examples, it is easy to guess that
$$u = n^3, n^3+1$$
are only $u$ we find. You can show this guess by substituting $u$ by $n^3+k$ for $k=0,1,2$, and checking that
$$f(u) = u^3 - n^3(u+1)^2$$
is increasing for $u \ge n^3+1$.
