Recognizing $x=1$ as an exact root is not something you "arrive at algebraically" so much as a search process through a discrete space of possibilities, followed by a computation to verify that it is a root. The algebraic solution formula for cubic equations expresses $1$ as one of three possible values of a complex expression involved nested imaginary square and cube roots and it is hard to recognize $1$ from looking at that. A numerical equation solver would find $1.00000....$ as an answer strongly hinting that $x=1$ is a solution.
Here is what Wolfram Alpha says about the real root of $x^3 + x = a$ using some form of the cubic formula:
$x = \left\{\frac{\sqrt[3]{9 a+\sqrt{3} \sqrt{27 a^2+4}}}{\sqrt[3]{2} 3^{2/3}}-\frac{\sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a+\sqrt{3} \sqrt{27 a^2+4}}}\right\}$
which for $a=2$ is the nice expression $x = \left\{\frac{\sqrt[3]{18+\sqrt{3} \sqrt{112}}}{\sqrt[3]{2} 3^{2/3}}-\frac{\sqrt[3]{\frac{2}{3}}}{\sqrt[3]{18+\sqrt{3} \sqrt{112}}}\right\}$
Imagine here a computer and calculator-equipped generation of students solving algebra homework by machine and submitting those as answers... (By the way, there are factors of 2 and 3 in the solution formula that Mathematica treats as additional cube root extractions, which are not part of the cubic formula in its usual presentation. This is wrong from the Galois-theoretic point of view where solving a cubic needs only one square root operation and one cube root of something built from the square root. It might be necessary in a computer algebra system where Sqrt and CubeRoot are interpreted as unique functions and not abstract roots whose placement into the complex plane involves some choices.)
There are two facts behind the simplicity of this problem that are useful to understand:
Because the equation is $x^3 + x - 2=0$ with a first coefficient of $1$ and a last coefficient of $-2$ one knows that any rational solution is an integer dividing $-2$ and $x=\pm 1$ are the easiest cases to check. I think this principle is called the "rational root theorem".
Once you have a root $x=a$ then $(x-a)$ is a factor of the given degree 3 polynomial. I think this is called "factor theorem" or "remainder theorem" for polynomials (sorry for not knowing the terminology but certainly the Wikipedia will have it, under polynomials). This implies the preceding principle about divisibility. Anyway, $(x^3 + x - 2)/(x-1)$ is a quadratic polynomial and there are several ways to see that it is positive everywhere for real values of $x$.
If you know calculus you can check that $f(x) = x^3 + x -2$ is increasing as a function of $x$ so that $f(x)=2$ can have only one real solution.