MHB Cardano's method of solving cubics

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The discussion focuses on solving the cubic equation \(x^3 - 13x + 12 = 0\) using Cardano's method. The participants explore the calculations for \(u\) and \(v\), which are derived from the equation's coefficients, and how these relate to the roots of the cubic. The correct application of Cardano's formula reveals that the roots of the equation are \(x = 3\), \(x = 1\), and \(x = -4\). There is also a discussion about the nature of cube roots and the potential for multiple solutions. Ultimately, the conversation emphasizes the importance of careful calculation and understanding of complex roots in cubic equations.
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Solve using cardan's method.

$$x^3-13x+12=0$$

$$x=v+u$$

$$3uv=-p=13$$

$$v^3+u^3=-q=-12$$

$$27v^6+324v^3-13=0$$

$$v^3=\frac{-324\pm\sqrt{324^2-27*4*13}}{54}$$

Please solve for x.I know I am asking for too much,but seems like I am not able to get the desired answers even though nothing seems wrong with the method.

I would be so grateful towards anybody who helps.

For the moderator: please change * to x,I don't know how to do that.
Thanks
 
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Cardano's formula states that $x = u+v$ is a solution to $x^{3} +px+q =0$ where $u = \Big(-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3} \right)^{3}} \Big)^{\frac{1}{3}}$

and $v = \Big(-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3} \right)^{3}} \Big)^{\frac{1}{3}}$.

So for $x^{3}-13x+12 = 0$ we have $u = \Big(-6 + \sqrt{36 -\frac{2917}{27}} \Big)^{\frac{1}{3}} = \Big(-6 + \sqrt{-\frac{1225}{27}} \Big)^{\frac{1}{3}}$

$= \Big(-6 + \frac{35 i}{3 \sqrt{3}} \Big)^{\frac{1}{3}}$

which according to Wolfram Alpha equals $\Big( \frac{1}{216} \left( 9 + 5 \sqrt{3}i \right)^{3} \Big)^{\frac{1}{3}} = \frac{1}{6} \left(9 + 5 \sqrt{3} i \right)$

And $v$ is the complex conjugate of $u$, that is $\frac{1}{6} \left(9 - 5 \sqrt{3} i \right)$.

So $ x = u+v = \frac{18}{6} = 3$
 
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Random Variable said:
Cardano's formula states that $x = u+v$ is a solution to $x^{3} +px+q =0$ where $u = \Big(-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3} \right)^{3}} \Big)^{\frac{1}{3}}$

and $v = \Big(-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3} \right)^{3}} \Big)^{\frac{1}{3}}$.

But if I do it the other way,i.e,thinking that v has 6 values,won't I get 6(v+u) s or is 3 of them surely going to repeat

So for $x^{3}-13x+12 = 0$ we have $u = \Big(-6 + \sqrt{36 -\frac{2917}{27}} \Big)^{\frac{1}{3}} = \Big(-6 + \sqrt{-\frac{1225}{27}} \Big)^{\frac{1}{3}}$

$= \Big(-6 + \frac{35 i}{3 \sqrt{3}} \Big)^{\frac{1}{3}}$

which according to Wolfram Alpha equals $\Big( \frac{1}{216} \left( 9 + 5 \sqrt{3}i \right)^{3} \Big)^{\frac{1}{3}} = \frac{1}{6} \left(9 + 5 \sqrt{3} i \right)$

Is it the only possible cube root?

And $v$ is the complex conjugate of $u$, that is $\frac{1}{6} \left(9 - 5 \sqrt{3} i \right)$.

How can you be sure?

So $ x = u+v = \frac{18}{6} =
3$

What does wolframalpha say about the other cube roots?...and what about the other roots?please show me how you would those too...

Thanks a lot,but I am wondering what happened when I did it.Why is my discriminant not negative like yours?Why don't I get anyone of the desired roots?
 
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It is useful to remember that, given the values of u and v... $$ u= \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}\ v= \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}\ (1)$$

... if $r_{0}= 1$, $r_{1} = e^{i\ \frac{2}{3}\ \pi}$ and $r_{2} = e^{i\ \frac{4}{3}\ \pi}$ are the 'cubic roots' of 1, then the roots of the equation $x^{3} + p\ x + q=0$ are given by...

$$x_{0}= r_{0}\ u + r_{0}\ v\ ;\ x_{1}= r_{1}\ u + r_{2}\ v\ ;\ x_{3}= r_{2}\ u + r_{1}\ v\ (2)$$

With (1), (2) and a bit of patience You can find that the three roots of $x^{3} - 13\ x + 12 = 0$ are $x_{0}=3$, $x_{1}= 1$ and $x_{2}=-4$... Kind regards $\chi$ $\sigma$
 
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