Best way to solve $X^3-X^2-X-1=0$

Question

can anyone help me for this cubic equation ? can be solved without delta method?

$X^3-X^2-X-1=0$

(answer is $\sim 1.8393$)

Answer 1

Here's a solution that finds all three roots of the function

$$x^3-x^2-x-1=0$$ Substitute $y=x-\frac{1}{3}$: $$-\frac{4}{3}-y-(y+\frac{1}{3})^2+(y+\frac{1}{3})^3=0$$ Expanding this gives: $$y^3-\frac{4}{3}y-\frac{38}{27}=0$$ If $y=\frac{\lambda}{z}+z$ then $z=\frac{1}{2}\left(y+\sqrt{y^2-4\lambda}\right)$ $$-\frac{38}{27}-\frac{4}{3}\left(z+\frac{\lambda}{z}\right)+\left(z+\frac{\lambda}{z}\right)^3=0$$ Multiply both sides by $z^3$ $$z^6+z^4\left(3\lambda-\frac{4}{3}\right)-\frac{38z^3}{27}+z^2\left(3\lambda^2-\frac{4\lambda}{3}\right)+\lambda^3=0$$ Substitute $\lambda=\frac{4}{9}$ and $u=z^3$ $$u^2-\frac{38}{27}u+\frac{64}{729}=0$$ Choose one of the solutions $$u=\frac{1}{27}(19+3\sqrt{33})$$ Substitute back for $u=z^3$ $$z^3=\frac{1}{27}(19+3\sqrt{33})$$ Solving for $z$ $$z=\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}$$ $$z=-\frac{1}{3}\sqrt[3]{-19+3\sqrt{33}}$$ $$z=\frac{1}{3}(-1)^{2/3}\sqrt[3]{19+3\sqrt{33}}$$ Starting with the first solution, we substitute back for $z=\frac{y}{2}+\frac{1}{2}\sqrt{y^2-\frac{16}{9}}$ Solving this for $y$ gives $$y=\frac{4}{3\sqrt[3]{19+3\sqrt{33}}}+\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}$$ Substitute back for $y=x-\frac{1}{3}$ $$x=\frac{1}{3}+\frac{4}{3\sqrt[3]{19+3\sqrt{33}}}+\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}$$ This is the first solution for $x$, going on to the second solution $$z=-\frac{1}{3}\sqrt[3]{-19-3\sqrt{33}}$$ Substitute back for $z=\frac{y}{2}+\frac{1}{2}\sqrt{y^2-\frac{16}{9}}$ $$\frac{y}{2}+\frac{1}{2}\sqrt{y^2-\frac{16}{9}}=-\frac{1}{3}\sqrt[3]{-19-3\sqrt{33}}$$ Solve for $y$ $$y=\frac{4(-1)^{2/3}}{3\sqrt[3]{19+3\sqrt{33}}}-\frac{1}{3}\sqrt[3]{-19-3\sqrt{33}}$$ Substitute back for $y=x-\frac{1}{3}$ $$x=\frac{1}{3}+\frac{4(-1)^{2/3}}{3\sqrt[3]{19+3\sqrt{33}}}-\frac{1}{3}\sqrt[3]{-19-3\sqrt{33}}$$ Here's the second solution $$z=\frac{1}{3}(-1)^{2/3}\sqrt[3]{19+3\sqrt{33}}$$ For the last solution we substitute back for $z=\frac{y}{2}+\frac{1}{2}\sqrt{y^2-\frac{16}{9}}$ $$\frac{y}{2}+\frac{1}{2}\sqrt{y^2-\frac{16}{9}}=\frac{1}{3}(-1)^{2/3}\sqrt[3]{19+3\sqrt{33}}$$ Solve the equation for $y$ $$y=\frac{(-1)^{2/3}}{3}\sqrt[3]{19+3\sqrt{33}}-\frac{4}{3}\sqrt[3]{\frac{-1}{19+3\sqrt{33}}}$$ Substitute back for $x=y-\frac{1}{3}$ $$x=\frac{1}{3}+\frac{(-1)^{2/3}}{3}\sqrt[3]{19+3\sqrt{33}}-\frac{4}{3}\sqrt[3]{\frac{-1}{19+3\sqrt{33}}}$$ I have collected the 3 solutions that I found above here. $$x=\frac{1}{3}+\frac{4}{3\sqrt[3]{19+3\sqrt{33}}}+\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}\approx1.83928675521416113255$$ $$x=\frac{1}{3}+\frac{4(-1)^{2/3}}{3\sqrt[3]{19+3\sqrt{33}}}-\frac{1}{3}\sqrt[3]{-19-3\sqrt{33}}\approx-0.41964-0.60629i$$ $$x=\frac{1}{3}+\frac{(-1)^{2/3}}{3}\sqrt[3]{19+3\sqrt{33}}-\frac{4}{3}\sqrt[3]{\frac{-1}{19+3\sqrt{33}}}\approx-0.41964+0.60629i$$

Answer 2

@Darksonn 's answer certainly works, but if you only want the positive real solution, there is a "cubic formula" that is reasonably useful in this case.

Theorem: If the cubic equation $$X^3 + pX + q$$ ($p, q$ real) satisfies $$\frac{p^3}{27} + \frac{q^2}{4} \geq 0,$$ then a solution to the cubic equation is $$X = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{p^3}{27} + \frac{q^2}{4}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{p^3}{27} + \frac{q^2}{4}}}.$$

This is the Cardano-Tartaglia Formula.

In our case, we take @Darksonn 's $y$-substitution $\displaystyle y = x - \frac{1}{3}$ and get

$$y^3 - \frac{4}{3}y - \frac{38}{27} =0.$$

So $p = \displaystyle - \frac{4}{3}$, $q = \displaystyle - \frac{38}{27}$; one may check that in this case, $$ \frac{p^3}{27} + \frac{q^2}{4} = \frac{- 64 \cdot 4 + 38^2}{27^2 \cdot 4} = \frac{297}{27^2}, \quad -\frac{q}{2} = \frac{19}{27},$$ so that the answer becomes \begin{align} y &= \sqrt[3]{\frac{19}{27} + \sqrt{\frac{297}{27^2}}} + \sqrt[3]{\frac{19}{27} - \sqrt{\frac{297}{27^2}}}\\ & = \sqrt[3]{\frac{19 + \sqrt{297}}{27}} + \sqrt[3]{\frac{19 - \sqrt{297}}{27}} \\ &= \frac{1}{3} \left(\sqrt[3]{19 + 3\sqrt{33}} + \sqrt[3]{19 - 3\sqrt{33}} \right). \end{align}

Since $\displaystyle y = x - \frac{1}{3}$, $\displaystyle x = y + \frac{1}{3}$, and we get $$x = \frac{1}{3} + \frac{1}{3} \left(\sqrt[3]{19 + 3\sqrt{33}} + \sqrt[3]{19 - 3\sqrt{33}} \right).$$

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To see that our answer matches @Darksonn 's, note that the terms not already matching are $\sqrt[3]{19 - 3\sqrt{33}}$ on the one side and $\displaystyle \frac{4}{\sqrt[3]{19 + 3 \sqrt{33}}}$ on the other side. To see this equality, note that for all positive $a, b$, $$a - b = \frac{a^2 - b^2}{a + b}$$ and letting $a = 19$, $b = 3 \sqrt{33}$, we get $$19 - 3 \sqrt{33} = \frac{64}{19 + 3 \sqrt{33}}$$ Taking the cube root of both sides, we're done.

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If you wish to handle the case where the discriminant $$\frac{p^3}{27} + \frac{q^2}{4}$$ is negative, or to understand the shortcomings of the above formula, I strongly recommend looking at Lecture 4, "Equations of Degree Three and Four," in Fuchs and Tabachnikov, Mathematical Omnibus [Amer. Math. Soc., Providence, 2007]. It is a good "elementary" explanation of what is going on.

Yet other good methods are mentioned in the answers to the similar problem https://math.stackexchange.com/questions/612765/find-roots-of-the-cubic-equation-x3-x2-3-0-without-using-calculator?rq=1 .