Wolfram Alpha uses a scary-looking algebraic method to solve $x^{3}-x-1=0$ that I've never encountered before. It seems that every other online algebra calculator uses numerical methods, such as the Newton-Raphson method. As I don't have a premium subscription to WA, I cannot paste the step-by-step here, but I will painstakingly type out the main steps shown on the mobile app:
Solve for x over the real numbers: $x^{3}-x-1=0$
Change coordinates by substituting $x=y+\frac{λ}{y}$, where $λ$ is a constant value that will be determined later: $-1-y-\frac{λ}{y}+\left(y+\frac{λ}{y}\right)^{3}=0$
Multiply both sides by $y^3$ and collect in terms of $y$:
$y^{6}+y^{4}\left(3λ-1\right)-y^{3}+y^{2}\left(3λ^{2}-λ\right)+λ^{3}=0$
Substitute $λ=\frac{1}{3}$ and then $z=y^{3}$, yielding a quadratic equation in the variable $z$:
$z^{2}-z+\frac{1}{27}=0$
Find the positive solution to the quadratic equation and substitute back for $z=y^{3}$:
$y^{3}=\frac{1}{18}\left(9+\sqrt{69}\right)$
Taking cube roots gives $\frac{\sqrt[3]{9+\sqrt{69}}}{\sqrt[3]{2}\cdot3^{\frac{2}{3}}}$ times the third roots of unity (I won't write them here), substitute each value of $y$ into $x=y+\frac{1}{3y}$, bring each solution to a common denominator and simplify:
$x=\frac{2\sqrt[3]{\frac{3}{\sqrt{69}+9}}+\sqrt[3]{2\left(\sqrt{69}+9\right)}}{6^{\frac{2}{3}}}$
OR $x=\frac{1}{6}\left(2\left(-3\right)^{\frac{2}{3}}\sqrt[3]{\frac{2}{\sqrt{69}+9}}-2^{\frac{2}{3}}\sqrt[3]{-3\left(\sqrt{69}+9\right)}\right)$
OR $x=\frac{1}{6}\left(\left(-2\right)^{\frac{2}{3}}\sqrt[3]{-3\left(\sqrt{69}+9\right)}-2\left(3\right)^{\frac{2}{3}}\sqrt[3]{\frac{-2}{\sqrt{69}+9}}\right)$
First of all, what is this "changing coordinates" sorcery and why does it work? Second, as Anakin Skywalker once asked, is it possible to learn this power (as a Grade 12 student with a little more math knowledge than average) and what can I search for to learn more about it?
P.S. None of the math teachers in my school know anything about it.
As said in the comments, this is Cardano's method, which is quite old (renaissance innovation!).
You have the equation $x^3 - x -1= 0$. In the Cardano method (which was geometrically inspired, I believe) you set $x = u+v$ and then impose a condition on the $u$ and $v$ to simplify to a quadratic. This will work as there is no quadratic term (if there is one, we do a linear change of variables to get rid of it first)
So $$(u+v)^3 - (u+v) - 1= 0$$ which simplifies to
$$0 = u^3 + 3u^2 v + 3u v^2 + v^3 - (u+v)-1 = u^3 + v^3 + 3uv(u+v)-(u+v) -1 $$
which implies $$u^3 + v^3 - 1 = 0\tag{1}$$ if we assume $$3uv=1\tag{2}$$ so that the middle mixed terms cancel out. But that last condition on $uv$ means that we know $u^3v^3$ too; cube the relation on both sides and we have
$$27u^3v^3=1 \tag{3}$$ so we're in a situation where if we set $a=u^3$, $b=v^3$, we know that $$a+b=1 \text{ and } ab = \frac{1}{27}$$
which reduces to a quadratic (!) equation in $a$ or $b$:
$$a + \frac{1}{27a} = 1 \implies a^2 + \frac{1}{27} = a \implies a^2 - a + \frac{1}{27}=0\tag{4}$$
Now solve $4$, so you know $a$ (two options, usually) and thus $u^3$ and $u$ and $b$ too, hence $v^3$ and $v$ too and then $x = u+v$ is the looked for solution.
Quite a nifty find by old Cardano, eh? Of course he had to do it in roman numerals for extra credit, and without modern notations.
The method you describe, setting $x = y + \lambda \frac{1}{y}$, is doing the two first steps at once: writing the solution as sum of two terms that have a constant (to be determined) product. Note that you get the exact same quadratic.