Solving $ x^3 - 3x + \sqrt 2 = 0$ First of all, I'm going to use plain text for my formulas since I'm using a screen reading program, which makes it very difficult to read the highly non-standard format commonly used here, sorry about that.
Okay, suppose we have a cubic equation like this:
x^3 - 3*x + sqrt(2) = 0
This is a "depressed cubic" so it should be possible to solve directly.
According to Wikipedia, the rational root test should be able to suggest whether there might be any rational solutions for our equation.
However, this only works for rational solutions, which means that I can't find out whether there's a simple expression for one solution unless that solution is rational. Consequently, if I don't know any of the 3 solutions in advance, then I can't factor it out to get a quadratic.
One way of solving at least a part of the task would be finding a general procedure for taking cube roots. I tried to do that by simply inverting the operations of "a^3 + 3*a*b^2" or "3*a^2*b + b^3" but I realized that I always had to guess either "a" or "b" (or more precisely, one of its 3 possible values) and then the other one came out. If I made a wrong guess, then the intermediate squares didn't match and I didn't find anything. Therefore, it didn't help me in what I really wanted to get to.
One possible example where it does work is finding the cube root of 387*sqrt(5) - 326*sqrt(7):
I can divide 387 by 3, which makes 129. I subtract 3*3*5 and divide the result by 3, which gives me 28. I divide that by 7 which I took the square root of and I have 4, which is a square of 2 or -2. When I try it the other way round, I see that +2 doesn't work and -2 does, which means that the cube root is:
3*sqrt(5) - 2*sqrt(7).
So in this case I really got an expression for the cube root but I had to guess that "a" might be 3*sqrt(5) and I got the corresponding "b" then.
Now, to solve x^3 - 3*x + sqrt(2) = 0, the rational root test is unusable and therefore I'm supposed to sum two complex cube roots. But it definitely must be possible either to express each of them without using cube roots or to apply a different procedure of finding a solution which can be expressed without them.
Thank you a lot for any suggestions.
Petr
 A: $x=\sqrt 2$ is a root by observation. Factor that out and you are left with a quadratic.
How did I guess that? Well the $\sqrt 2$ on its own was matched with odd powers of $x$, so if I wrote $x=y\sqrt 2$ I could see that the $\sqrt 2$ would factor out. Indeed we get $$2\sqrt 2 y^3-3\sqrt 2 y+\sqrt 2=0$$ Then cancel out the $\sqrt 2$ to get $$2y^3-3y+1=0$$ to which you can apply the rational root test etc.
A: Hint:
You can easily rearrange the statement to: 
Plaintext: x^3+\sqrt{2} = 3 x
LaTeX: $x^3+\sqrt{2} = 3 x$
Thus easy to see that 
Plaintext: x = sqrt(2), x = -(1+sqrt(3))/sqrt(2), or x = (sqrt(3)-1)/sqrt(2)
LaTeX: $x = \sqrt{2}$, $x = -\frac{1+\sqrt{3}}{\sqrt2}$ or $x = \frac{1+\sqrt{3}}{\sqrt2}$
Solving should prove easy.
A: Not sure whether this is an answer to your question, but by inspection $\sqrt 2$ sqrt 2 is obviously a root. (As Mark Bennet explained I found this by observing that because of which terms the square root was multiplying, a change of variables absorbing the square root into the variable was possible.)
x^3-3x+sqrt 2= (x-sqrt 2)(x^2 +sqrt 2 x-1)
$$x^3-3x+\sqrt 2= (x-\sqrt 2)(x^2 +\sqrt 2 x-1)$$
