Solving a complex cubic equation

Question

I am trying to solve the following equation: $$ z^3 + z +1=0 $$

Attempt: I tried to factor out this equation to get a polynomial term, but none of the roots of the equation is trivial.

Answer 1

Indeed, none of the roots is trivial. In this case, your best option is probably to apply Cardano's formula. It gives a real root and two complex non-real roots.

Answer 2

Hint:

Like How to solve the cubic $x^3-3x+1=0$?,

let $z=a\cos t,a\ne0$ such that $\dfrac{a^3}4=\dfrac{-a}3\implies a^2=-\dfrac43\implies a=\pm\dfrac2{\sqrt3}i$

$$-1=a^3\cos^3t+a\cos t=\dfrac{a\cos3t}{-3}$$

$$\iff\cos3t=\dfrac3a$$

If we choose $a=-\dfrac{2i}{\sqrt3},\cos3t=\dfrac{3\sqrt3i}2,\sin3t=\pm\sqrt{1+\dfrac{27}4}$

$$e^{i3t}=i\left(\dfrac{3\sqrt3}2\pm\sqrt{1+\dfrac{27}4}\right)$$

Answer 3

Given your equation $x^3+x+1=0$, in standard form ($ax^3+bx^2+cx+d=0$) your equation has $$\qquad a=1\qquad b=0\qquad c=1\qquad d=1$$

The "common" cubic formula is

$$x=\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)+\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}+\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)-\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}-\frac{b}{3a}$$

and if you plug in the coefficients, the $b$ terms will drop out and you will have something a "little" simpler. You may encounter transient complex terms under the square root but WolframAlpha can deal with them. You can also give WolframAlpha a try at solving the equation directly as it does here yielding $1$ real and $2$ complex roots. I think the real root you are seeking is here.

Answer 4

Apply the cubic formula.

https://math.stackexchange.com/a/4868747/928654

We get

$$z_1=\sqrt[3]{-\frac12+\sqrt{\frac14+\frac{1}{27}}}+\sqrt[3]{-\frac12-\sqrt{\frac14+\frac{1}{27}}}=\sqrt[3]{-\frac12+\frac{\sqrt{31}}{6\sqrt3}}+\sqrt[3]{-\frac12-\frac{\sqrt{31}}{6\sqrt3}}$$

This is about $-0.6823$

if $i^2=-1$

$$z_2=(\frac{-1+i\sqrt3}{2})\sqrt[3]{-\frac12+\sqrt{\frac14+\frac{1}{27}}}+(\frac{-1-i\sqrt3}{2})\sqrt[3]{-\frac12-\sqrt{\frac14+\frac{1}{27}}}=(\frac{-1+i\sqrt3}{2})\sqrt[3]{-\frac12+\frac{\sqrt{31}}{6\sqrt3}}+(\frac{-1-i\sqrt3}{2})\sqrt[3]{-\frac12-\frac{\sqrt{31}}{6\sqrt3}}=-\frac12\sqrt[3]{-\frac12+\frac{\sqrt{31}}{6\sqrt3}}+\frac{i\sqrt3}{2}\sqrt[3]{-\frac12+\frac{\sqrt{31}}{6\sqrt3}}-\frac12\sqrt[3]{-\frac12-\frac{\sqrt{31}}{6\sqrt3}}-\frac{i\sqrt3}{2}\sqrt[3]{-\frac12-\frac{\sqrt{31}}{6\sqrt3}}=-\frac12(\sqrt[3]{-\frac12+\frac{\sqrt{31}}{6\sqrt3}}+\sqrt[3]{-\frac12-\frac{\sqrt{31}}{6\sqrt3}})+\frac{i\sqrt3}{2}(\sqrt[3]{-\frac12+\frac{\sqrt{31}}{6\sqrt3}}-\sqrt[3]{-\frac12-\frac{\sqrt{31}}{6\sqrt3}})$$

This is about $0.3412+i1.1615$

One interesting fact about nonreal solutions to polynomial equations over the real numbers is that they always come in complex conjugate pairs.

$$z_3=-\frac12(\sqrt[3]{-\frac12+\frac{\sqrt{31}}{6\sqrt3}}+\sqrt[3]{-\frac12-\frac{\sqrt{31}}{6\sqrt3}})-\frac{i\sqrt3}{2}(\sqrt[3]{-\frac12+\frac{\sqrt{31}}{6\sqrt3}}-\sqrt[3]{-\frac12-\frac{\sqrt{31}}{6\sqrt3}})$$

This is about $0.3412+i1.1615$

Answer 5

Substitute $x=\frac2{\sqrt3}\sinh t$ to rewrite the equation $x^3+x+1=0 $ as

$$4\sinh^3t+3\sinh t + \frac{3\sqrt3}2=0$$

Comparing with the identity $4\sinh^3t+3\sinh t = \sinh3t$ results in $\sinh 3t =- \frac{3\sqrt3}2$, or $t = -\frac13\sinh^{-1} \frac{3\sqrt3}2$. Thus, one real root of the cubic equation is $$x_0= -\frac2{\sqrt3}\sinh \left( \frac13\sinh^{-1} \frac{3\sqrt3}2\right) \approx -0.6823 $$

Then, factorize the cubic equation as $$x^3+x+1=(x-x_0)(x^2+x_0 x-\frac1{x_0})$$ and the quadratic factor gives a pair of complex roots $$x_{1,2}=-\frac{x_0}2 \pm \frac i2 \sqrt{1-\frac3{x_0}}$$