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Solve $x^3 = 3px + 2q$ when $q^2 < p^3$ by the substitution $x = 2 \sqrt{p} \cos(\theta)$. The solutions are given by $x = 2 \sqrt{p} \cos(\frac{1}{3}(\phi + 2k\pi))$ for $k = 0,1,2$ where $\cos\phi = \frac{q}{p\sqrt{p}}$

I have made the subsitution to get:

$$8p^{\frac{3}{2}} (\cos(\theta))^3 = 6p^{\frac{3}{2}}\cos(\theta) + 2q $$ $$ q = 4p^{\frac{3}{2}}(\cos(\theta))^3 - 3p^{\frac{3}{2}}\cos(\theta)$$

$ q = p^{\frac{3}{2}} \cos(3 \theta)$ (this last identity I got from a previous question)

$$\frac{q}{p^{\frac{3}{2}}} = \cos(3 \theta)$$

I am unsure how to proceed from here, any hints?

Michael Hardy
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yhu
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1 Answers1

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You have done the hard work,

let $\displaystyle\cos\phi=\frac q{p^{\frac32}}$ which is possible iff $\displaystyle-1\le \frac q{p^{\frac32}}\le 1\iff 0\le \frac{q^2}{p^3}\le1$

If $p>0$ and $ q^2\le p^3$ (which is given)

So, we have $\displaystyle\cos3\theta=\cos\phi\implies 3\theta=2n\pi\pm\phi$ where $n$ is any integer

$\displaystyle\implies \theta=\frac{2n\pi\pm\phi}3 $ where $n$ can assume any three in-congruent values $\pmod3$ when either '+' or '-' sign is considered as there are exactly three roots

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lab bhattacharjee
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