# Solving a real cubic using trigonometric methods

The problem is to figure out how to solve a real cubic equation of the form $x^3 + px + q = 0$ using trigonometry. The first step is to prove the identity

$$4\cos^3 \theta - 3\cos \theta - \cos 3\theta = 0$$

I can do this using de Moivre's formula as per the hint provided. I just equate the real parts of the expression for $\left(\cos \theta + i\sin \theta\right)^3$ with $\cos 3\theta$, replace $\sin^2 \theta$ with $1 - \cos^2 \theta$ and everything cancels out.

The next step is to set $y = ax$ in $x^3 + px + q = 0$ and show that there is a value of $a$ for which the equation then becomes $4y^3 - 3y - q_1 = 0$. I'm not sure exactly how to go about doing this. It seems like the only sensible thing to do is solve for $x$ in terms of $y$ to get $x = y/a$ and then plug that into the original equation. Doing so yields

$$\left(\frac{y}{a}\right)^3 + p\frac{y}{a} + q = \frac{1}{a^3}y^3 + \frac{p}{a}y + q = 0.$$

We can see then that $a$ must satisfy $a = 4^{-1/3}$ and $a = -p/3$. This uniquely determines $p$ as $p = -3/4^{1/3}$ so I know that this can't be the right thing to do.

What does it mean to set $y = ax$?

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It's not quite right to say that the equation becomes $4 y^3 - 3 y - q_1 = 0$, rather that it becomes equivalent to $4 y^3 - 3 y - q_1 = 0$, namely it is a (nonzero) constant multiple of that equation. So multiply your $\frac{1}{a^3} y^3 + \frac{p}{a} y + q$ by something that makes the coefficient of $y^3$ equal to 4, and then see what $a$ must be...
I must have woken up on the slow and stupid side of the bed this morning. $\frac{1}{a^3}c = 4 \Rightarrow c = 4a^3 \Rightarrow -3 = 4a^2p \Rightarrow a=\frac{i}{2}\sqrt{\frac{3}{p}}$ but the problem says that we'll use only reals and the next step would then be to find an angle $\theta$ such that $cos 3\theta = -\frac{i}{2}\left(\frac{3}{p}\right)^{\frac{3}{2}}q$ so either $p$ or $q$ would have to have a factor of $i$ but this is a real cubic. I must be missing something very obvious here but I can't find it. –  knucklebumpler Apr 3 '11 at 23:15