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I've been trying to solve this for hours and all found was the real solution by Cardano"s formula.

I vaguely remember that if $\alpha$ is a root of a complex number, the other roots are $\omega \alpha$ and $\omega ^2 \alpha$ where $\omega = - \frac{1}{2}+i \frac{\sqrt{3}}{2}$, $\omega ^2 = - \frac{1}{2}-i \frac{\sqrt{3}}{2}$.

The problem is, I don't know how to apply this to find my roots(if applicable at all).

Straightly put, how can I find other roots of a cubic given one root?(This time it happens to be around -0.75, for the real root). Given a single root, what is the relationship between others? Any formula to compute the remaining solutions at all?

NoChance
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John Trail
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    You can do real polynomial division to get the other roots. – Peter Jan 17 '16 at 22:41
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    If you can find one root, use either synthetic or long division to factor $$y^3 - y/3 + 25/27 = (y - \text{root})p(y)$$ where $p$ is a degree $2$ polynomial. –  Jan 17 '16 at 22:41
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    The numbers $\alpha, \omega\alpha, \omega^2\alpha$ are the three cube roots of $\alpha^3$. They are also roots of the equation $y^3 - \alpha^3=0$, but you do not have that form of the cubic equation: yours has the additional term $-\frac13y$. – David K Jan 17 '16 at 22:47
  • finding $p(y)$ doesn't seem to work; I set $(y+0.75488)(y^2+ay+b)=y^3-\frac{1}{3}+\frac{25}{27}$ and find $a,b$ but there's no such $a,b$(-0.75488) was my root). I guess it's supposed to be complex? the $a,b$? But if the above method doesn't work by comparing coeffiicients, I don't know what will. Division sounds very very tedious to me...is that my only option? – John Trail Jan 17 '16 at 22:51
  • When you have a real irrational root and two complex roots, Cardano's gives all three. Read more carefully. https://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method – Will Jagy Jan 17 '16 at 23:05
  • I've been looking through the Cardano section but I don't really see your statement in it... the "irrational real root and 2 complex roots exists, then Cardano gives all three". The bit I am stuck(in terms of English maybe) is where it says "the fact uv is real implies that one is obtained by multiplying the first by..." I mean, which "one" is multiplied to give what "first?" – John Trail Jan 17 '16 at 23:12
  • it seems to instruct me to multiply my real root by $-\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$ to get the 2 other complex solutions but this does not give me the right answer..... – John Trail Jan 17 '16 at 23:15

2 Answers2

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By $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a+\omega b+\omega^{2} c)(a+\omega^{2} b+\omega c)$;

and replacing $a, b, c$ by $x, -u, -v$ respectively,

$x^{3}-3uvx-(u^{3}+v^{3})=0 \implies x_{k}=u\, \omega^{k}+ v\, \omega^{2k}$ for $k=0,1,2$.

$u, v$ are known as resolvents.

Let your reduced cubic be $y^{3}-3py-2q=0$.

(Will substitute $p=\frac{1}{9}, q=-\frac{25}{54}$ later.)

Now $uv=p$ and $2q=u^{3}+v^{3}$.

$\therefore 2q=u^{3}+\left( \frac{p}{u} \right)^{3}$

$\implies u^{6}-2qu^{3}-p^{3}=0$

$\implies u^{3}=q\pm \sqrt{q^{2}-p^{3}}$

$\implies u=\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}}$

$\displaystyle{ \therefore v=\frac{p}{\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}}} =\sqrt[3]{q\mp \sqrt{q^{2}-p^{3}}} }$

$\therefore y=\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}}+\sqrt[3]{q\mp \sqrt{q^{2}-p^{3}}}$.

Note that $u, v$ are conjugates and symmetrical in roles.

By keeping the upper case,

$y=\sqrt[3]{q+\sqrt{q^{2}-p^{3}}}+\sqrt[3]{q-\sqrt{q^{2}-p^{3}}}$ is one of the root.

The other two roots are $\omega u+\omega^{2} v$ and $\omega^{2} u+\omega v$.

In this case, $u+v=-\frac{1}{3} \left( \sqrt[3]{\frac{25+3\sqrt{69}}{2}}+\sqrt[3]{\frac{25-3\sqrt{69}}{2}} \right) $.

Careful manipulation in $\omega$ enables us to find the rest (complex roots).

P.S.: In case of the discriminant $\Delta = q^{2}-p^{3} <0$, there are 3 distinct real roots involving combination of $\cos(\frac{1}{3} \cos^{-1} (.))$ and not discussed here.

Ng Chung Tak
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In theory, the answer to your question of finding the other two roots of a cubic equation knowing one of them is simply: it is enough to divide the cubic polynomial $ax^3+bx^2+cx+d$ by the linear factor $(x-x_0)$, where $x_0$ is the root you know, so you get a quadratic trinomial you can solve (this goes without saying that the value $x_0$ must be exact and not approximate)

However, in practice this is not always easily feasible and when $x_0$ has a complicated expression the above mentioned division could be hard

In the particular case of your equation, changing $3y$ by $x$ you have the trinomial $x^3-3x+25$ and the real root $x_1$ $$x_1=-{\sqrt[3]{\frac{2}{25-3\sqrt{69}}}}-{\sqrt[3]{\frac{25-3\sqrt{69}}{2}}}$$ Put by confort $$M=\frac{2}{25-3\sqrt{69}}\space\text{so}\space x_1=-\left(\sqrt[3]M+\sqrt[3]{\frac1M}\right)$$ the quotient above mentioned is equal to $$x^2-\left(\sqrt[3]M+\sqrt[3]{\frac 1M}\right)x+\left(\left(\sqrt[3]M+\sqrt[3]{\frac 1M}\right)^2-3\right)$$ Hence, solving this quadratic gives $$2x_{2,3}=\sqrt[3]M+\sqrt[3]{\frac 1M}\pm\sqrt{12-3\left(\sqrt[3]M+\sqrt[3]\frac 1M\right)^2}$$ This gives the other two roots that are not real. But the calculation of its values, being direct, can be tedious.

Piquito
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