solve $t^5+t+1=0$ I honestly don't have any idea at all on how to solve this.
I am asked to find solutions under $\mathbb{R},\mathbb{Q},\mathbb{C}$ respectively but this seems impossible to solve without a computer.
Does anyone know how to go about this? Thank you
 A: HINT:
$$t^5+t+1=0\Longleftrightarrow$$
$$\left(t^2+t+1\right)\left(t^3-t^2+1\right)=0$$
Split into two equations:


*

*Use the 'abc-formula' to find the solution to: 
$$t^2+t+1=0$$

*$$t^3-t^2+1=0\Longleftrightarrow$$

Eliminate the quadratic term by substituting $x=t-\frac{1}{3}$:

$$1-\left(x+\frac{1}{3}\right)^2+\left(x+\frac{1}{3}\right)^3=0\Longleftrightarrow$$
$$x^3-\frac{x}{3}+\frac{25}{27}=0\Longleftrightarrow$$

If $x=y+\frac{\lambda}{y}$ then $y=\frac{1}{2}\left(x+\sqrt{x^2-4\lambda}\right)$:

$$25+\frac{1}{3}\left(-y-\frac{\lambda}{y}\right)+\left(y+\frac{\lambda}{y}\right)^3=0\Longleftrightarrow$$

Multiply both sides by $y^3$ and collect in terms of $y$;
Substitute $\lambda=\frac{1}{9}$ and then $z=y^3$, yielding a quadratic equation in the variable $z$:

$$z^2+\frac{25z}{27}+\frac{1}{729}=0$$
A: $$t^5+t+1=(t^2+t+1)(t^3-t^2+1)=0$$
The equation $t^2+t+1=0$ can easily be solved. The equation $t^3-t^2+1=0$ can be solved either numerically or with the (complicated) formula for cubic equations.
A: For $\mathbb Q$ one can use the rational root theorem. 
This gives that the only $1$ and $-1$ can be rational roots, but they clearly aren't. 
