Solve $x^7-5x^4-x^3+4x+1=0$ for $x$
Consider the identity
$(x^4-4x-1)^2=x^8-8x^5-2x^4+16x^2+8x+1$
Differentiating both sides w.r.t. $x$, we get,
$x^7-5x^4-x^3+4x+1=(x^4-4x-1)(x^3-1)$
Now, the equation becomes,
$(x^4-4x-1)(x^3-1)=0$
$\implies x=1,\omega, \omega^2$ (where $\omega$ is a non real cube root of unity)
or
$x^4-4x-1=0$
$\implies x^{4}+2x^{2}+1=2x^{2}+4x+2$
$\implies (x^{2}+1)^{2}=2(x+1)^{2}$
$\implies \{x^{2}+1+\sqrt{2} (x+1)\}\cdot \{x^{2}+1-\sqrt{2} (x+1)\}=0$
$\implies x=\dfrac{-\sqrt{2} \pm i\sqrt{2+4\sqrt{2}}}{2}$
or
$x=\dfrac{\sqrt{2} \pm \sqrt{4\sqrt{2}-2}}{2}$
$\therefore x=1,\omega,\omega^2,\dfrac{-\sqrt{2} \pm i\sqrt{2+4\sqrt{2}}}{2},\dfrac{\sqrt{2} \pm \sqrt{4\sqrt{2}-2}}{2}$
you will get $$(x-1)(x^2+x+1)(x^4-4x-1)=0$$ and you can solve your problem