Solve the equation $$\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1$$
$$\Leftrightarrow \sqrt{x^3+1}=\frac{2x^3+x^2+1}{4x-1}$$
$$\Leftrightarrow \sqrt{x^3+1}-\left(x+1\right)=\frac{2x^3+x^2+1}{4x-1}-\left(x+1\right)$$
$$\Leftrightarrow \frac{x^3+1-\left(x+1\right)^2}{\sqrt{x^3+1}+x+1}=\frac{2x^3-3x^2-3x+2}{4x-1}$$
$$\Leftrightarrow \frac{x^3-x^2-2x}{\sqrt{x^3+1}+x+1}-\frac{2x^3-3x^2-3x+2}{4x-1}=0$$
$$\Leftrightarrow \frac{x\left(x-2\right)\left(x+1\right)}{\sqrt{x^3+1}+x+1}-\frac{\left(x+1\right)\left(x-2\right)\left(2x-1\right)}{4x-1}=0$$
$$\Leftrightarrow \left(x-2\right)\left(x+1\right)\left(\frac{x}{\sqrt{x^3+1}+x+1}-\frac{2x-1}{4x-1}\right)=0$$
The equation $$\frac{x}{\sqrt{x^3+1}+x+1}-\frac{2x-1}{4x-1}=0$$ have a solution but it's very ugly and i can't solve this. Help me, thanks
after squaring rearranging and factorizing we get $$-(x-2) x (x+1) \left(4 x^3-8 x^2+9 x-4\right)=0$$