Best way to solve $X^3-X^2-X-1=0$ can anyone help me for this cubic equation ?
can be solved without delta method?
$X^3-X^2-X-1=0$
(answer is $\sim 1.8393$)
 A: @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
$$x = \frac{1}{3} + \frac{1}{3} \left(\sqrt[3]{19 + 3\sqrt{33}} +  \sqrt[3]{19 - 3\sqrt{33}} \right).$$

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.

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 .
