How can I solve the equation $x^3-x-1 = 0$? Can someone give me a hint on how can I solve the equation
$$x^3 - x - 1 =0?$$
Thank you!
 A: I like to apply Cardano's method. 
It can be used immediately (without a change of variable) because you have to find the roots of a depressed cubic.
Let $x=u+v$, where $u$ and $v$ are two complex variables that I'll define later.
Then your equation can be written
$$(u+v)^3-(u+v)-1=0\;.$$
By expanding, you find
$$u^3+3u^2v+3uv^2+v^3-u-v-1=0\;,$$
which can be written
$$u^3+v^3-1+(u+v)(3uv-1)=0\;.$$
Let's try to find $u$ and $v$ by imposing these two conditions:
$$u^3+v^3=1$$
$$uv=\dfrac 13$$
Then $x=u+v$ would clearly be a solution of your equation.
The second condition implies that $u^3v^3=\dfrac 1{27}$.
We know the sum $S$ and the product $P$ of the two numbers $U=u^3$ and $V=v^3$. 
It's a well-known fact that $U$ and $V$ are the roots of the quadratic equation $X^2-SX+P=0$. 
You can simply expand $(X-U)(X-V)$ if you're not convinced. 
In our case, we have to solve for
$$X^2-X+\dfrac 1{27}=0\;.$$
When the real numbers $U$ and $V$ are found ($U$ can be either of the two roots), you can then find three possible complex values for $u$. If you want the real solution of your initial equation, then you simply take the real cubic root of $U$, and then $v$ is uniquely defined by $v=\dfrac 1{3u}$ (or you can also take the real cubic root of $V$ for $v$).
I found:
$$x=\sqrt[3]{\dfrac{9+\sqrt{69}}{18}}+\sqrt[3]{\dfrac{9-\sqrt{69}}{18}}\approx 1.325$$
The other cubic roots of $U$ are $ju$ and $j^2u$, where $j=e^{2i\pi/3}$ and $u$ is the real cubic root of $U$.
A: HINT:
Let $z=a\cos y, z^3-z=a^3\cos^3y-a\cos y$ where $a>0$
Comparing with $\cos3y=4\cos^3y-3\cos y$
$$\dfrac43=\dfrac{a^3}a\implies a=\dfrac2{\sqrt3}=\dfrac1{\cos30^\circ}$$
Now $\cos3x=\cos3A\implies3x=m360^\circ\pm3A$ where $m$ is any integer
$x=A,A+120^\circ,A+240^\circ$
