# Derive the polynomial

Given that the solutions to a cubic equation using Cardano's method are $$x_1, x_2, x_3=\sqrt[3]{-\frac{260}{9}i\sqrt{3}-21}+\sqrt[3]{\frac{260}{9}i\sqrt{3}-21}-3$$ derive the cubic polynomial and its factors using algebraic methods only, i.e. without using trigonometric functions.

My initial thought is I need to find the cube roots of the nested radicals which I think could be done with DeMoivre’s formula. But since that is not permitted I am unsure of where to start to get the cube roots.

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Do you mean that $x_1 = x_2 = x_3$? – Antonio Vargas Mar 15 '13 at 22:17
You have only been given one solution. From that one cannot recover the equation. But if we assume the cubic has rational coefficients, you can find the other solutions, and then find the coefficients using information about the connection between the solutions and the coefficients. – André Nicolas Mar 15 '13 at 22:17

A way to get to the cube roots of the nested radicals in the question, thus getting to the three separate roots "buried" in the Cardano formula above, without using DeMoivre's formula to find the cube roots of the nested radicals is, first set

$x=-21$ and $y=-\frac{260}{9}i\sqrt{3}$ and (without going into how the equation below was derived) then find any real roots of the following equation

$$\frac{(-64a^9+(48x)a^6+((15(x)^2)-3(3y)^2)a^3+(x)^3)}{-64} = 0$$

This expands to $$a^9 +\frac{63}{4}a^6 - \frac{74215}{64}a^3 +\frac{9261}{64} = 0$$

and as luck would have it has three rational roots $a_1=3;$ $a_2=\frac{1}{2}$; $a_3=-\frac{7}{2}$. Next solve the equations

$$a_1^3+3a_1b_1^2 = -21$$ $$a_2^3+3a_2b_2^2 = -21$$ $$a_3^3+3a_3b_3^2 = -21$$ (where $-21$ is the value set as $x$) for $b_1$, $b_2$ and $b_3$, or $$b_1=\pm\frac{4i\sqrt{3}}{3}$$ $$b_2=\pm\frac{13i\sqrt{3}}{6}$$ $$b_3=\pm\frac{5i\sqrt{3}}{6}$$ So the three cube roots of $$\sqrt[3]{-\frac{260}{9}i\sqrt{3}-21}$$ are $a$+$b$ or $$a_1+b_1=-\frac{4i\sqrt{3}}{3}+3$$ $$a_2+b_2=\frac{13i\sqrt{3}}{6}+\frac{1}{2}$$ $$a_3+b_3=-\frac{5i\sqrt{3}}{6}-\frac{7}{2}$$

The same method would be used to derive the cube roots of $$\sqrt[3]{\frac{260}{9}i\sqrt{3}-21}$$

Summarize all the calculations including the cube roots of of $$\sqrt[3]{\frac{260}{9}i\sqrt{3}-21}$$, (which we did not calculate here), to get the following roots of the equation

$$x_1=\left(-\frac{4i\sqrt{3}}{3}+3\right) +\left(\frac{4i\sqrt{3}}{3}+3\right) -3 = 3$$

$$x_2=\left(\frac{13i\sqrt{3}}{6}+\frac{1}{2}\right)+\left(-\frac{13i\sqrt{3}}{6}+\frac{1}{2}\right) -3 = -2$$

$$x_3=\left(-\frac{5i\sqrt{3}}{6}-\frac{7}{2}\right)+\left(\frac{5i\sqrt{3}}{6}-\frac{7}{2}\right)-3=-10$$ Multiplying the factors $$(x-3)(x+2)(x+10)=0$$ equals $$x^3+9x^2-16x-60=0$$ the polynomial the question is seeking to derive.

Indeed, Cardano's formula for this cubic equation is the same as that presented in the original question and the three cube roots of the nested radicals in the equation are in the solutions found.

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