# How can I solve the equation $x^3-x-1 = 0$? [closed]

Can someone give me a hint on how can I solve the equation $$x^3 - x - 1 =0?$$

Thank you!

## closed as off-topic by user91500, Shailesh, Claude Leibovici, JMP, JunivenFeb 23 '17 at 10:54

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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 $U$.

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

$x=A,A+120^\circ,A+240^\circ$

• I don't understand why do you compare with $cos 3 y = 4 cos^3 y -3 cosy.$ Can you give me more details? Thank you! – g.pomegranate Feb 23 '17 at 10:42
• @g.pomegranate, As we have $x^3, x$ only – lab bhattacharjee Feb 23 '17 at 10:56