We need to find only the positive root for the equation: $$x^3-3x^2-x=\sqrt{2}$$ I start to think firstly in substitution like taking $x=u+1$ this helps me only in eliminating the quadratic term to obtain the following equation: $$u^3-4u-3-\sqrt{2}=0$$ Now I can continue to obtain my root?
Is it useful to take substitution $x=2v+1$ which will give me the new equation $v^3-v=\frac{\sqrt{2}-3}{8}$?
We seek a "rational" root – in a quadratic extension of the rationals.
Our sought root will have the form $a+b\sqrt2$ with $a,b$ rational. Since the equation has a unit leading coefficient and all other coefficients have the form $p+q\sqrt2$ with $p,q$ integers, our rational values $a,b$ must give an integer Euclidean norm:
$N=(a+b\sqrt2)(a-b\sqrt2)=a^2-2b^2\in\mathbb{Z}$
and thus the rational $a,b$ are themselves integers.
We plug $x=a+b\sqrt2$ into the equation, expand the powers and collect the rational terms (those not containing $\sqrt2$) to obtain a necessary condition for the root:
$\underset{\text{from }(a+b\sqrt2)^3}{(a^3+6ab^2)}-3\underset{\text{from }(a+b\sqrt2)^2}{(a^2+2b^2)}-a=0\tag{1}$
The constant term in the equation for $x$ has norm
$0^2-2×1_2=-2,$
so the norm $N=a^2-2b^2$ of our root must divide this. Plugging $2b^2=a^2-N$ into (1) then gives
$4a^3-6a^2-(3N+1)a+3N=0, N\in\{\pm1,\pm2\}$
If we put in $N=1$ we get
$4a^3-6a^2-4a+3=0,$
which fails to give an integer root for $a$. The trial that works turns out to be:
$N=2\implies 4a^3-6a^2-7a+6=0\implies a=2.$
So we have candidate roots $2+\sqrt2,2-\sqrt2$ which both have rational term $2$ and norm $2^2-2=2$. Out of these two, $2+\sqrt2$ holds and thus we have our root
$x=2+\sqrt2.$
The remaining roots are obtainable by factoring this out and solving the residual quadratic equation by standard methods. The discriminant of the quadratic equation will turn out negative, so these roots are complex conjugates.
I suggest an alternative solution than the ones in the comments) by using a very helpful formula $$a^3+b^3+c^3 -3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \tag{1}$$ Let us define $a,b$ such that $$\begin{align} &a^3+b^3 = -(3+\sqrt{2})\\ &3ab = -4 \end{align}$$ It’s easy to solve the quadratic equation $w^2 +(3+\sqrt{2})w + \left(4/3 \right)^3=0$ for $w\in\{a^3,b^3\}$: $$(a^3,b^3) = \left( \frac{-(3+\sqrt{2}) \color{red}{\pm} \sqrt{(3+\sqrt{2})^2 -4^4/3^3 }}{2} \right)$$ From $(1)$, we deduce the unique solution is $$u = -(a+b) $$
The resulting expression for $u$ cannot be simplified algebraically, but direct numerical evaluation indicates that it matches $u=x-1=1+\sqrt2$, where $x=2+\sqrt2$ is obtained by a direct reduction procedure (see here).
Squaring the equation and rearranging it into a sixth-degree polynomial gives:
$$x^6-6x^5+7x^4+6x^3+x^2-2=0$$
Now, let's try to factor it. Linear factors won't help because the polynomial has no rational roots. And cubic factors won't help because that's what we started with. So that leaves quadratic times quartic:
$$(x^2 + Ax + B)(x^4 + Cx^3 + Dx^2 + Ex + F) = 0$$ $$x^6 + (A + C)x^5 + (B + D + AC)x^4 + (E + AD + BC)x^3 + (F + AE + BD)x^2 + (AF + BE)x + BF = 0$$
Matching up the coefficients of this polynomial with the original gives us a system of 6 equations for 6 coefficients. Solving it is a bit tedious, but it turns out that $(A=-4, B=2, C=-2, D=-3, E=-2, F=-1)$ works. IOW,
$$(x^2 - 4x + 2)(x^4 - 2x^3 - 3x^2 - 2x - 1) = 0$$
For the first factor, the quadratic formula gives $x = 2 \pm \sqrt{2}$. It turns out that $x = 2 - \sqrt{2}$ fails to satisfy the original equation, but $\boxed{x = 2 + \sqrt{2}}$ works.
And then dividing the original cubic polynomial by its known factor gives $\frac{x^3-3x^2-x-\sqrt{2}}{x - (2 + \sqrt{2})} = x^2 + (\sqrt{2} - 1)x + (\sqrt{2} - 1)$. Applying the quadratic formula gives:
$$x = \frac{1 - \sqrt{2} \pm \sqrt{7 - 6\sqrt{2}}}{2}$$
But those aren't real numbers, so they're not the root you're looking for.
Let $t=x-\sqrt2$ or $x=t+\sqrt2$ and then the equation becomes $$ t^3+3(\sqrt2-1)t^2+(5-6\sqrt2)t-6=0. $$ It is easy to see that $t=2$ is a root. So $$ t^3+3(\sqrt2-1)t^2+(5-6\sqrt2)t-6=(t-2)(t^2+(-1+3\sqrt2)t+3)=0$$ which can be solved easily.
