Solve for $x$ : $x^4 = 5(x-1)(x^2 - x +1)$

Question

Solve for $x$ : $x^4 = 5(x-1)(x^2 - x +1)$

I am having difficulty factoring the whole equation so that I can equate each factor to $0$ one by one and get the roots accordingly.

Answer 1

Hint: let $x=y+1$, expand, regroup and collect, then the equation becomes:

$$y^4-y^3+y^2-y+1=0$$

Multiplying by $y+1 \ne 0$ gives $y^5 + 1 = 0$ having as solutions the $5^{th}$ roots of $-1$.

Answer 2

Divide both sides by $x^4$ and change variable to $y = \frac{x-1}{x^2}$, we have

$$\begin{align}1 = 5y(1-y) \iff & 4y(y-1) = -\frac45 \iff (2y-1)^2 = \frac15\\ \implies & y = \frac12\left(1 \pm \frac{1}{\sqrt{5}}\right) = \frac{\sqrt{5}\pm 1}{2\sqrt{5}} = \frac{2}{5 \mp \sqrt{5}} \end{align} $$ Substitute $y$ back by $\frac{x-1}{x^2}$, we get

$$ x^2 - \frac{5\mp\sqrt{5}}{2} (x - 1) = 0 \iff \left(x - \frac{5\mp\sqrt{5}}{4}\right)^2 = \frac{5\mp\sqrt{5}}{2} - \left(\frac{5\mp\sqrt{5}}{4}\right)^2 = -\frac{5 \pm \sqrt{5}}{8} $$ This leads to $\displaystyle\;x = \frac{5 - \epsilon \sqrt{5} \pm i \sqrt{2(5 + \epsilon \sqrt{5})}}{4}$ where $\epsilon = \pm 1$.

If one insists on factoring the equation first, one can treat $x^2$ and $(x-1)$ as two seperate units and rearrange them:

$$\begin{align} & x^4 = 5(x-1)(x^2 - x + 1)\\ \iff & x^4 - 5x^2(x-1) + 5(x-1)^2 = 0\\ \iff & \left(x^2 - \frac52(x-1)\right)^2 = \left(\frac{25}{4} - 5\right)(x-1)^2 = \frac{5}{4}(x-1)^2\\ \iff & \left(x^2 - \frac{5 + \sqrt{5}}{2}(x-1)\right)\left(x^2 - \frac{5 - \sqrt{5}}{2}(x-1)\right) = 0 \end{align} $$