<s>[INST] Solve the equation $\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1$ 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 
 [/INST] After squaring and rearranging we have:
$$4x^6-12x^5+9x^4+3x^3-14x^2+8x=0$$
Since there is no constant term, $x=0$ is a solution. Using the rational root theorem, $p$ and $q$ have to be coprime. Since the only factors of $4$ and $8$ are powers of $2$, then the only solutions are when either $1$ is in the numerator or $1$ is the denominator:
$$x = ±\frac{1}{8}, ±\frac{1}{4}, ±\frac{1}{2}, ±1, ±2, ±4, ±8$$
Verifying these solutions by substitution, only $x=-1$ and $x=2$ are possible. We can factorise them (for instance, by polynomial/synthetic division) to get the cubic in Dr. Sonnhard Graubner's answer.
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