Solving a algebraic equation I am to solve the following equation third order equation
$x^3+x^2+x+1=0$
What I've tried so far is writing the equation as
$ x \cdot (x^2+x+1)+1=0$ 
but that didn't lead anywhere. How do I solve this without using a computer?

Two approaches to finding a root: (We can confirm a real root exists, by the rational root theorem).
$$x^3+x^2+x+1=0$$


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*We can recognize that $x = -1$ is a zero by using a quick check at $x = \pm 1$ (again, by the rational root theorem, those are possible zeros). Then we know that since $x = -1$ is a zero, $(x - (-1)) = (x + 1)$ is a factor, and we can just use polynomial long division to obtain the remaining factor of $(x^2 + 1)$.

*We can "group" the equation as follows: $$x^3+x^2+x+1=0 \iff x^2(x + 1) + (x + 1) = (x+1)(x^2 + 1) = 0$$
Finally, we know that the remaining factor in $(x+1)(x^2 + 1)=0$, namely $(x^2 + 1),\;$ has no real root (and hence, cannot be factored) by evaluating the discriminant $$\Delta = b^2 - 4ac = 0 - 4 = -4 < 0$$
(Recall the discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by the $\color{blue}{\bf radicand}$ of the quadratic formula $$\frac{-b \pm  \sqrt{\color{blue}{\bf b^2 - 4ac}}}{2a}$$
