# 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

• Yikes! Please try both to use LaTeX for mathematics writing in this site (directions in the FAQ section), no matter what screen you use (it's either you or us suffering...), and please do use here and there some lines spacing! – DonAntonio Jun 7 '13 at 9:59
• @DonAntonio He's using screen reader accessibility software, not a strange screen; I'm not sure if I understand what you mean when you say "no matter what screen you use". Can the OP read things rendered in LaTeX at all? – Sharkos Jun 7 '13 at 10:06
• @Sharkos, I don't know, but he said it is difficult for him to read the "highly non-standard"(??) format used here, and for us is highly difficult to read unlatexed mathematical sutff, so it is either he or we that are going to have problems reading stuff... – DonAntonio Jun 7 '13 at 12:00
• It is highly non standard by the standards of the internet; I don't know what the effect of the HTML on a screen reader is? – Sharkos Jun 7 '13 at 12:10
• By Cardano's formula, if $x^3 + cx = d$ then $$x = \sqrt [3] {\frac d2 + \sqrt{\frac{d^2}{4} + \frac{c^3}{27}}} + \sqrt [3] {\frac d2 - \sqrt{\frac{d^2}{4} + \frac{c^3}{27}}}$$ and by substituting $c = -3$ and $d = \sqrt{2}$, and then with some simplifying and then evaluating, we obtain that... – Mr Pie Mar 6 '18 at 4:19

$x=\sqrt 2$ 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.

• To put the formulae without LaTeX x=sqrt2 is a root. To find this without guessing, note that since x appears to odd powers, the substitution x=y*sqrt2 gives an equation with integer coefficients, namely 2y^3-3y+1=0, and the rational root theorem then provides y=1 as an easy solution. – Mark Bennet Jun 7 '13 at 10:12
• Thanks to all of you for the answers. One final question: Is there a rule which would let me know in advance whether the solutions for a given cubic can be reduced further or not? I'm now thinking of the frequently quoted example of "x^3 - 3*x + 1 = 0" whose solutions don't seem to be reducible but I have no way of knowing this before trying to solve it. – Petr Jun 7 '13 at 15:19

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.

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)

$$x^3-3x+\sqrt 2= (x-\sqrt 2)(x^2 +\sqrt 2 x-1)$$