Polynomials – Solving Cubic Equation with Real Roots

cubicspolynomialsroots-of-cubics

Consider the equation,

$$(x+b)^3 = a x^2\tag1$$

with $a\ne 0$, and real roots $(x_1$, $x_2$, $x_3)$. Show that,

\begin{eqnarray}\sqrt[3]{x_1} &+& \ \sqrt[3]{x_2} &+ &\ \sqrt[3]{x_3} &=& \sqrt[3]{a}\\
\sqrt[3]{\frac{1}{x_1}}&+&\sqrt[3]{\frac{1}{x_2}}&+&\sqrt[3]{\frac{1}{x_3}}&=&\ 0
\end{eqnarray}

Notes:

  1. Ramanujan had a series of equalities with cubic roots of trigonometric functions. The ones of second type are of the form
    $$\sqrt[3]{\frac{\cos \frac{2 \pi}{7}}{\cos{ \frac{4 \pi }{7}} }}+\sqrt[3]{\frac{\cos \frac{4 \pi}{7}}{\cos{ \frac{6 \pi }{7}} }}+\sqrt[3]{\frac{\cos \frac{6 \pi}{7}}{\cos{ \frac{2 \pi }{7}} }}= \sqrt[3]{-7}$$
    since the above quotients of cosines are the roots of $x^3 + 4 x^2 + 3 x -1= (x-1)^3 – (-7) x^2$ One has similar results considering $\cos$ of multiples of $\frac{2 \pi}{9}$.

  2. There are general formulas for $(\sqrt[3]{x_1} + \sqrt[3]{x_2} + \sqrt[3]{x_3})^3$ being the root of a cubic polynomial with a unique real root, obtained in an algebraic way from the initial polynomial. So the above is on the beaten path. However, I did not find this case posted; it seems cute, and easy to prove, so maybe useful as a reference.

Any feedback would be appreciated!

$\bf{Added:}$

  1. We could also consider the cubic form
    $$(x+y)^3 – x^2 z + f x z ( x+y+ \frac{f^2}{27} z)$$
    Again, if the roots $x_i$ in $x$ are real and distinct then
    \begin{eqnarray}
    \sqrt[3]{x_1} + \ \sqrt[3]{x_2} + \ \sqrt[3]{x_3}=\sqrt[3]{z}
    \end{eqnarray}

  2. One could try something similar for say $(x+b)^5 – a x^k$, $1\le k \le 4$, however the resulting equations do not have all roots real, since after taking the $5$-th root we get a polynomial with gaps in coefficients.

Best Answer

The OP asks for any feedback. In fact, the simple-looking cubic,

$$(x+\beta)^3 = \alpha x^2\tag1$$

has a LOT of interesting properties, being connected to Ramanujan's theta function, the Klein quartic curve, and septic equations. First, some minor transformations. Let $x = 1/y,\; \alpha=n^2+13n+49,$ and $\beta=1$ to get,

$$y^3+3y^2-(n^2+13n+46)y+1 = 0\tag{2a}$$

This retains the two properties mentioned by the OP, plus some additional ones. Given the Ramanujan theta function $f(a,b)$, Dedekind eta function $\eta(\tau)$, and $q=e^{2\pi i\tau}$, define,

\begin{align} a &= -q^{61/168}\,\frac{f(-q,-q^6)}{f(-q^2)}\\ b &= \; q^{13/168}\;\frac{f(-q^2,-q^5)}{f(-q^2)}\\ c &= q^{-11/168}\,\frac{f(-q^3,-q^4)}{f(-q^2)}\\ n &= \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4\\ \psi &= \left(\frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(7\tau)}\right)^4 \end{align}

such that $(a,b,c)$ obey the Klein quartic curve,

$$a^3b+b^3c+c^3a = 0$$

Then the roots $y_k$ of the cubic have the form,

\begin{align} y_1 &= -(a^3b)^3\,\psi\\ y_2 &= -(b^3c)^3\,\psi\\ y_3 &= -(c^3a)^3\,\psi \end{align}

which explains why the roots of,

$$y^3+3y^2-(n^2+13n+46)y+1 = 0\tag{2b}$$

even for any $n$ have,

$$y_1^{1/3} + y_2^{1/3} + y_3^{1/3} = 0$$

since they are just obeying the Klein quartic curve. However, if we take 7th roots instead,

$$z=-(y_1^3\,y_2)^{1/7}-(y_2^3\,y_3)^{1/7}-(y_3^3\,y_1)^{1/7}\tag3$$

then $z$ is a solution in radicals to the solvable septic,

$$z^7 + 14 z^4 - 21 z^3 - 14 (46 + 13 n + n^2) z^2 - 7 (249 + 114 n + 18 n^2 + n^3) z - (1348 + 854 n + 207 n^2 + 23 n^3 + n^4) = 0$$

For example, let $n=-6$, then the septic,

$$z^7 + 14 z^4 - 21 z^3 - 56 z^2 + 21 z - 4 = 0$$

is solvable by the cubic,

$$y^3 + 3 y^2 - 4 y + 1 = 0$$

using the relation $(3)$ and the roots $y_k$ correctly ordered. There's more, but this is enough. And to think all these inter-connections came from the simple structure $(x+\beta)^3 = \alpha x^2.$

Addendum: Since the OP asked about the special functions $(a,b,c)$ of order 7 (more on these functions in this MO post), they are generalizations of the Roger-Ramanujan identities of order 5. Define,

\begin{align} u &=\, q^{11/60}\,\frac{f(-q,-q^4)}{f(-q)}\; =\; q^{11/60}\prod_{n=1}^\infty \frac1{\big(1-q^{5n-2}\big)\big(1-q^{5n-3}\big)}\\ v &= q^{-1/60}\,\frac{f(-q^2,-q^3)}{f(-q)} = q^{-1/60}\prod_{n=1}^\infty \frac1{\big(1-q^{5n-1}\big)\big(1-q^{5n-4}\big)} \end{align}

which now is related to the quadratic,

$$x^2-(m+11)x-1=0$$

with roots $(x_1, x_2) = (-u^{10}m^5,\;v^{10}m^5)$ and $m = \left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6.$

Radicals

The eta quotients $(m,n,\psi)$ are radicals when $\tau = \sqrt{-d}$, which implies the special functions $(u,v)$ and $(a,b,c)$ are radicals are well. For example, let $d=1$, then,

\begin{align} m &= \left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6 = 5^3\left(\frac{1+\sqrt5}{2}\right)^3\\ n &= \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4 =\frac{28\times7^{3/4}\sqrt2}{\left(\sqrt{5-\sqrt7}-\sqrt{-7+3\sqrt7}\right)^2} \\ \psi &= \left(\frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(7\tau)}\right)^4 =\frac{7^{3}\sqrt{14}}{\left(\sqrt{5-\sqrt7}-\sqrt{-7+3\sqrt7}\right)^4} \\ \end{align}

Substituting these into the quadratic and cubic, then solving for their roots, one can then solve for $(u,v)$ and $(a,b,c)$ in terms of radicals.

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