I'm given a logarithm problem which is $$\log_{2} (x^3+1)-2\log_{2}x=\log_{2}(x^2-x+1)-2$$
I'm stuck in the step of $x^4-5x^3+x^2-4=0$
By many times of trial and error, I got $(x^2-4x-4)(x^2-x+1)=0$
My question, is there any standard way to factorise the equation without trials and errors? Thanks in advance.
Hint: the quartic can be avoided altogether by noting that $x^3+1=(x+1)(x^2-x+1)\,$. Given that $x^2-x+1 \gt 0$ on $\mathbb{R}$ it follows that $x^3+1$ and $x+1$ have the same sign, which must be positive for the $\log_2$ to be defined. Then the equation simplifies to:
$$ \require{cancel} \log_{2} (x+1)+\cancel{\log_{2} (x^2-x+1)}-2\log_{2}x=\cancel{\log_{2}(x^2-x+1)}-2 \\ \iff\quad \log_2{}\frac{x+1}{x^2}=-2 \quad\iff\quad \frac{x+1}{x^2}=\frac{1}{4} $$