Solve $\sqrt{x^2+1}-\sqrt{4x^4-4x^2+2}=2x^3-x-1$ 
Could you please help me solve for $x$ in
  $$\sqrt{x^2+1}-\sqrt{4x^4-4x^2+2}=2x^3-x-1.$$


I tried this way. But I could not solve further. Please help me.
$$(\sqrt{x^2+1}-\sqrt{4x^4-4x^2+2})^2=(2x^3-x-1)^2$$
 A: As $x$ increases:
  $\sqrt{4x^4-4x^2+2} = \sqrt{(2x^2-1)^2+1}$ decreases if $x \in [0,\frac12\sqrt2]$ and increases if $x \in [\frac12\sqrt2,-\infty)$.
  Symmetrically $\sqrt{4x^4-4x^2+2}$ decreases if $x \in (-\infty,-\frac12\sqrt2]$ and increases if $x \in [-\frac12\sqrt2,0]$.
Therefore $\sqrt{4x^4-4x^2+2} \le \sqrt{2}$ if $|x| \le \frac12\sqrt2$.
If $x \in [0,\frac12\sqrt2]$:
  $-(2x^3-x-1) = 1-x·(2x^2-1) \ge 1$.
If $x \in [-\frac12\sqrt2,0]$.
  $-(2x^3-x-1) = (1-x)·(\frac12(2x+1)^2+\frac12) \ge \frac12$.
If $|x| \le \frac12\sqrt2$:
  $\sqrt{4x^4-4x^2+2} \le \sqrt2 < 1+\frac12 \le 1-(2x^3-x-1)$.
If $x < -\frac12\sqrt2$:
  $\sqrt{4x^4-4x^2+2} \le 2x^2 < 1-(2x^3-x-1)$ because $(x+1)·(2x^2-1) < 1$.
Therefore $\sqrt{4x^4-4x^2+2} < \sqrt{x^2+1}-(2x^3-x-1)$ if $x \le \frac12\sqrt2$.
If $x > \frac12\sqrt2$:
  $\frac{d}{dx}(\sqrt{x^2+1}-(2x^3-x-1)) = \frac{x}{\sqrt{x^2+1}} - 6x^2+1 < 0$.
Therefore $\sqrt{x^2+1}-(2x^3-x-1) = \sqrt{4x^4-4x^2+2}$ for at most one real $x$, and indeed equality holds when $x = 1$.
