Q.Solve : $x^4 - x^3 + x^2 - x + 1 = 0$

This is a reciprocal (palindromic) equation; divide by $x^2$ and substitute $y=x+\dfrac1x$ to reduce it to a quadratic, then solve for $x$ from each value of $y$.

1. Equation: $x^4-x^3+x^2-x+1=0$. The coefficients are palindromic ($1,-1,1,-1,1$), so it is a reciprocal equation of even degree.
2. Since $x=0$ is not a root, divide throughout by $x^2$: $x^2-x+1-\dfrac1x+\dfrac{1}{x^2}=0 \;\Rightarrow\; \left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac1x\right)+1=0$
3. Let $y=x+\dfrac1x$. Then $x^2+\dfrac{1}{x^2}=y^2-2$.
4. Substituting: $(y^2-2)-y+1=0 \Rightarrow y^2-y-1=0$.
5. Solve for $y$: $y = \dfrac{1\pm\sqrt{1+4}}{2}=\dfrac{1\pm\sqrt5}{2}$.
6. Case $y_1=\dfrac{1+\sqrt5}{2}$: solve $x+\dfrac1x=y_1 \Rightarrow x^2-y_1x+1=0$. $x = \dfrac{y_1\pm\sqrt{y_1^2-4}}{2}$. Now $y_1^2=\dfrac{(1+\sqrt5)^2}{4}=\dfrac{6+2\sqrt5}{4}=\dfrac{3+\sqrt5}{2}$, so $y_1^2-4=\dfrac{3+\sqrt5-8}{2}=\dfrac{\sqrt5-5}{2}<0$ — complex roots. $x = \dfrac{1+\sqrt5}{4}\pm\dfrac{i}{2}\sqrt{\dfrac{5-\sqrt5}{2}}$.
7. Case $y_2=\dfrac{1-\sqrt5}{2}$: similarly $x^2-y_2x+1=0$, with $y_2^2=\dfrac{3-\sqrt5}{2}$ and $y_2^2-4=\dfrac{3-\sqrt5-8}{2}=\dfrac{-5-\sqrt5}{2}<0$. $x = \dfrac{1-\sqrt5}{4}\pm\dfrac{i}{2}\sqrt{\dfrac{5+\sqrt5}{2}}$. …