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Could anyone please help and clarify how this particular cubic equation

$$4p^3-4p-1=0$$

could be solved?

I am familiar with the approach of solving quadratic equations through quadratic formula and discriminant, but I am a bit confused about how to approach above cubic equation.

Quanto
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Joker312
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    You can use Cardano's method – DatBoi Nov 13 '20 at 12:09
  • The cubic equation has a rich history; see https://en.wikipedia.org/wiki/Cubic_equation for more than you may want to know. If you're looking for exact formula along the lines of $(-b\pm\sqrt{b^2-4ac})/(2a)$ for the quadratic equation, you'll have to plow through the part on Cardano. If approximate solutions are good enough, there are various techniques that'll hone in on answers. – Barry Cipra Nov 13 '20 at 12:09
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    Question has surely been asked and answered here many times. A search for "Cardano" should find several instances. – Gerry Myerson Nov 13 '20 at 12:11
  • https://math.stackexchange.com/questions/378366/a-method-for-solving-cubic-equation – vvg Nov 13 '20 at 12:12
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    The Cardano method however is quite complicated and requires complex roots (in the case of $3$ real roots). So, usually they are either solved numerically or with geometric methods. – Peter Nov 13 '20 at 12:42

3 Answers3

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$4p^3-4p-1=0$

$p^3-p-\frac 14=0$

Substitute $p = q + \frac 1{3q}$, expand the above.

You will get an equation of the form $q^3 + \frac a{q^3} + b = 0$, which can be transformed into a quadratic by substituting $q^3 = r$ and multiplying throughout by $r \ (\ne 0)$. Solve for $r$, retaining any complex roots and reverse all the substitutions.

More explanation: what you presented is essentially a "depressed cubic" as it lacks a square term. Any general cubic can be transformed into a depressed cubic via a linear substitution, following which it can be solved by further substitutions and transformations as above. It's just tedious and not commonly done.

(Also, I presented the technique as applied in your particular case. The general method can be elucidated by using symbolic variables in place of the coefficients, but that would make my answer longer. I am working on the assumption that you just want to know how to solve this particular cubic).

Deepak
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It is quite simple to follow the steps given here.

The key point is to first compute $\Delta$; in your case it is $592$ so we know that there are three ral roots. Now, we know that we have three real roots. For this case, the trigonometric method gives nice formulae.

Using the method, the roots are given by $$x_k=\frac{2 }{\sqrt{3}}\cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(\frac{3 \sqrt{3}}{8}\right)\right)\right)\qquad \text{for} \qquad k=0,1,2$$

Claude Leibovici
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Substitute $p=\frac2{\sqrt3}\cos\theta$ into $4p^3-4p-1=0 $ to get

$$4\cos^3\theta - 3\cos\theta =\cos3\theta=\frac{3\sqrt{3}}8$$ which leads to $\theta = \frac13 \cos^{-1}\frac{3\sqrt{3}}8+\frac{2\pi k}3$ and the solutions

$$p= \frac{2 }{\sqrt{3}}\cos \left(\frac13 \cos^{-1}\frac{3\sqrt{3}}8 +\frac{2\pi k}3 \right), \>\>\>k=0,1,2$$

Quanto
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