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\begin{center}
\vskip 1cm{\LARGE\bf Ramanujan Cubic Polynomials\\
\vskip .1in
of the Second Kind}
\vskip 1cm
\large
Roman Witu{\l}a \\
Institute of Mathematics \\
Silesian University of Technology \\
Kaszubska 23 \\
Gliwice 44-100 \\
Poland \\
\href{mailto:roman.witula@polsl.pl}{\tt roman.witula@polsl.pl} \\
\end{center}


\vskip .2in

\begin{abstract}
We present generalizations of some identities discussed earlier
by Shevelev. Moreover, we introduce Ramanujan cubic
polynomials of the second kind (RCP2). This new type of cubic
polynomial is closely related to the Ramanujan cubic
polynomials (RCP) defined by Shevelev.
We also give many fundamental properties
of RCP2's.
\end{abstract}


\vskip .2in






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\section{Shevelev type identities}

V.~Shevelev \cite{Shevelev2009} gave a trigonometric equality of the form
\begin{equation}\label{1}
\sqrt{\frac{\cos\frac{2\pi}{5}}{\cos\frac{\pi}{5}}}+
\sqrt{\frac{\cos\frac{\pi}{5}}{\cos\frac{2\pi}{5}}}=\sqrt{5}.
\end{equation}

The theorem, given below, shows that (\ref{1})
is a special case of a large class
of identities for Fibonacci numbers~$F_n$:

\begin{theorem}
We have
\begin{equation}\label{2}
\sqrt[r]{\frac{F_{n-1}\varphi ^{r-1}}{\varphi ^{n-1}-F_n}}+
\sqrt[r]{\frac{\varphi ^{n-1}-F_n}{F_{n-1}\varphi ^{r-1}}}=\sqrt{5},\qquad n,r\in\mathbb{N},
\end{equation}
where $\varphi$ denotes the golden ratio $\bigl(\varphi
=\frac{1+\sqrt{5}}{2}\bigr)$.
\end{theorem}

\begin{proof}
We note that (\ref{2}) is
a~consequence of the following identities
\begin{equation*}
\varphi +\varphi ^{-1}=\sqrt{5}
\end{equation*}
and
\begin{equation*}
\varphi ^n=F_n\varphi +F_{n-1}
\end{equation*}
(which is proved by a simple induction)
or, equivalently,
\begin{equation*}
\varphi ^r=\frac{F_{n-1}\varphi ^{r-1}}{\varphi ^{n-1}-F_n}.
\end{equation*}
\end{proof}

In the next theorem we present 
identities (\ref{2}) for the general Fibonacci sequences
\begin{equation}\label{3}
\left\{\begin{array}{l}
F_0^{\ast}=0,\quad F_1^{\ast}=1, \\
F_{n+1}^{\ast}=\lambda _1F_n^{\ast}+\lambda _2F_{n-1}^{\ast},\quad n\in\mathbb{N},
\end{array}\right.
\end{equation}
where $\lambda _1,\lambda _2\in\mathbb{C}$, $\lambda _1^2+4\lambda _2\neq 0$, $\lambda_2\neq 0$.

Let $x_1,x_2$ be two roots of the characteristic equation
\begin{equation*}
x^2-\lambda _1x-\lambda _2=0.
\end{equation*}
We note that $x_1\neq x_2$. Then we have

\begin{theorem}
The following identities hold:
\begin{equation}\label{4}
x_l^n=F_n^{\ast}x_l+\lambda _2F_{n-1}^{\ast},
\end{equation}
\begin{equation}\label{5}
\frac{\lambda _2F_{n-1}^{\ast}}{x_l^{n-1}-F_n^{\ast}}-\frac{x_l^{n-1}-
F_n^{\ast}}{F_{n-1}^{\ast}}=\lambda _1,
\end{equation}
\begin{equation}\label{6}
\frac{x_l^n-\lambda _2F_{n-1}^{\ast}}{\sqrt{\lambda _2}F_n^{\ast}}-
\frac{\sqrt{\lambda _2}F_n^{\ast}}{x_l^n-\lambda _2F_{n-1}^{\ast}}=
\frac{\lambda _1}{\sqrt{\lambda _2}},
\end{equation}
\begin{equation}\label{7}
\bigg(\frac{\lambda
_2F_{n-1}^{\ast}}{x_1}+F_n^{\ast}\bigg)^{\frac{k}{n-1}}+
\bigg(\frac{\lambda _2F_{n-1}^{\ast}}{x_2}+F_n^{\ast}\bigg)
^{\frac{k}{n-1}}= \lambda _1F_k^{\ast}+2\lambda _2F_{k-1}^{\ast},
\end{equation}
for any $l=1,2$ and $k,n\in\mathbb{N}$.
\end{theorem}

\begin{proof}

(\ref{4}). Equality (\ref{4}) can be proven by induction with
respect to $n\in\mathbb{N}$.
\bigskip

\noindent
(\ref{5}). From (\ref{4}) we get
\begin{equation*}
x_l=\frac{\lambda _2F_{n-1}^{\ast}}{x_l^{n-1}-F_n^{\ast}}.
\end{equation*}
Next, we note that
\begin{equation}\label{8}
x_l^2-\lambda _1x_l-\lambda _2=0\quad\Leftrightarrow\quad x_l-\lambda _2x_l^{-1}=
\lambda _1.
\end{equation}
\bigskip

\noindent
(\ref{6}). From (\ref{4}) we have
\begin{equation*}
x_l=\frac{x_l^n-\lambda _2F_{n-1}^{\ast}}{F_n^{\ast}}.
\end{equation*}
Hence, by (\ref{8}) we obtain (\ref{6}).
\bigskip

\noindent
(\ref{7}). From (\ref{4}) we receive ($l=1,2$):
\begin{equation*}
x_l^{n-1}=\frac{\lambda _2F_{n-1}^{\ast}}{x_l}+F_n^{\ast}
\quad\Rightarrow\quad
x_l=\sqrt[n-1]{\frac{\lambda _2F_{n-1}^{\ast}}{x_l}+F_n^{\ast}}
\end{equation*}
(the last one holds for the respective value of $(n-1)$-th root of
the number $\frac{\lambda _2F_{n-1}^{\ast}}{x_l}+F_n^{\ast}$), which
implies the identity
\begin{equation*}
\bigg(\frac{\lambda _2F_{n-1}^{\ast}}{x_1}+F_n^{\ast}\bigg)^{\frac{k}{n-1}}+
\bigg(\frac{\lambda _2F_{n-1}^{\ast}}{x_2}+F_n^{\ast}\bigg)^{\frac{k}{n-1}}=
x_1^k + x_2^k
%={} \\
{}\stackrel{\mbox{\rm (\ref{4})}}{=}
\lambda_1 F_k^{\ast} + 2\lambda_2 F_{k-1}^{\ast}.
\end{equation*}
\end{proof}

\begin{corollary}
For any $k,n\in \mathbb{N}$ the following identity holds:
\begin{multline}
\Big( F_{n+1} + \frac{F_n}{\varphi} \Big)^{k/n} +
\Big( F_{n+1} - \frac{F_n}{\varphi - 1} \Big)^{k/n} = {}\\
{}=
\Big( F_{n-1} + \varphi\, F_n \Big)^{k/n} +
\Big( F_{n+2} - \varphi^2\, F_n \Big)^{k/n} =
F_k + 2\, F_{k-1} = L_k,
\end{multline}
where $L_k$ denotes the $k$-th Lucas number.
\end{corollary}

\begin{remark}
Identities, similar to those discussed in the previous theorem, can
be generated for the elements of linear recurrence equations of
any order. See in particular the relations defining the so-called
quasi-Fibonacci
numbers~\cite{WitulaSlota-A7,WitulaSlota-A7s,WitulaSlota-A11,WitulaSlota-A13}.
\end{remark}

\begin{remark}
Shevelev's intention in the 
paper~\cite{Shevelev2009} was, it seems, to investigate the
sum
\begin{equation*}
\sqrt{\bigg|\frac{x_1}{x_2}\bigg|}+\sqrt{\bigg|\frac{x_2}{x_1}\bigg|},
\end{equation*}
where $x_1,x_2\in\mathbb{R}$ are roots of the polynomial
\begin{equation*}
x^2-\lambda_1 x-\lambda _2
\end{equation*}
and $x_1x_2<0$. Then we have
\begin{multline}\label{9}
\sqrt{\bigg|\frac{x_1}{x_2}\bigg|}+\sqrt{\bigg|\frac{x_2}{x_1}\bigg|}=
\frac{|x_1|+|x_2|}{\sqrt{|x_1x_2|}}=\sqrt{\frac{(|x_1|+|x_2|)^2}{|x_1x_2|}}={} \\
{}=\sqrt{\frac{(x_1+x_2)^2+2|x_1x_2|-2x_1x_2}{|x_1x_2|}}=
\sqrt{\frac{\lambda _1^2+4\lambda _2}{\lambda _2}}.
\end{multline}
In the particular case of
\begin{equation*}
x^2+x-1=\bigg( x+2\cos\frac{\pi}{5}\bigg)\bigg(
x-2\cos\frac{2\pi}{5}\bigg),
\end{equation*}
the identity (\ref{1}) follows from (\ref{9}).
\end{remark}

The extension of sums (\ref{9}) to sums for
roots of a given cubic polynomial is described in the next section.


\begin{remark}
We note that (see formula~(\ref{7})):
\begin{equation}
\lambda_1\, F_{k}^{\ast} + 2\, \lambda_2\, F_{k-1}^{\ast} =
F_{k+1}^{\ast} + \lambda_2\, F_{k-1}^{\ast} =
L_{k}^{\ast},
\end{equation}
where $L_{k}^{\ast}$ denotes the generalized Lucas sequence
\begin{equation}\label{3l}
\left\{\begin{array}{l}
L_0^{\ast}=2,\quad L_1^{\ast}=\lambda_1, \\
L_{n+1}^{\ast}=\lambda_1\, L_n^{\ast} + \lambda_2\, L_{n-1}^{\ast},\qquad n\in\mathbb{N}.
\end{array}\right.
\end{equation}
\end{remark}

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\section{Cubic Shevelev sums}



Let us assume that $\xi_{1}$, $\xi_{2}$, $\xi_{3}$
are complex roots of the following polynomial with complex coefficients
$$
f(z) := z^{3} + p\,z^{2} + q\,z + r.
$$
The symbols $\sqrt[3]{\xi_{1}}$, $\sqrt[3]{\xi_{2}}$,
$\sqrt[3]{\xi_{3}}$ will denote any of the third complex roots of
the numbers $\xi_{1}$, $\xi_{2}$ and $\xi_{3}$, respectively (only
in the case that $\xi_{1}$, $\xi_{2}$ and $\xi_{3}$ are real
numbers we will assume that $\sqrt[3]{\xi_{1}}$,
$\sqrt[3]{\xi_{2}}$ and $\sqrt[3]{\xi_{3}}$ also denote the
respective real roots).

Let us set
$$
A :=
\left(
\sqrt[3]{\xi_{1}} + \sqrt[3]{\xi_{2}} + \sqrt[3]{\xi_{3}}
\right)^{3}
$$
and
$$
B := \left( \sqrt[3]{\xi_{1}}\, \sqrt[3]{\xi_{2}} +
\sqrt[3]{\xi_{1}}\, \sqrt[3]{\xi_{3}} +
\sqrt[3]{\xi_{2}}\, \sqrt[3]{\xi_{3}}  \right)^{3}.
$$
Thus, the numbers
$$
\sqrt[3]{\xi_{1}} + \sqrt[3]{\xi_{2}} + \sqrt[3]{\xi_{3}}
\qquad  \mbox{ and } \qquad
\sqrt[3]{\xi_{1}}\, \sqrt[3]{\xi_{2}} +
\sqrt[3]{\xi_{1}}\, \sqrt[3]{\xi_{3}} +
\sqrt[3]{\xi_{2}}\, \sqrt[3]{\xi_{3}}
$$
belong to the sets of the third complex roots of $A$ and~$B$,
respectively, which, for the conciseness of notation, will be
denoted by the symbols $\sqrt[3]{A}$ and~$\sqrt[3]{B}$,
respectively. In other words, we have
$$
\sqrt[3]{\xi_{1}} + \sqrt[3]{\xi_{2}} + \sqrt[3]{\xi_{3}} \in \sqrt[3]{A}
$$
and
$$
\sqrt[3]{\xi_{1}}\, \sqrt[3]{\xi_{2}} +
\sqrt[3]{\xi_{1}}\, \sqrt[3]{\xi_{3}} +
\sqrt[3]{\xi_{2}}\, \sqrt[3]{\xi_{3}} \in \sqrt[3]{B} .
$$


Then we can deduce the relation
\begin{equation}\label{10}
27 A B = (A+p-3\sqrt[3]{r})^3.
\end{equation}
We note that
\begin{multline*}
\xi _1^{1/3}\xi _2^{1/3}\xi _3^{1/3}\bigg(\frac{\xi _1^{1/3}}{\xi _2^{1/3}}+
\frac{\xi _2^{1/3}}{\xi _1^{1/3}}+\frac{\xi _2^{1/3}}{\xi _3^{1/3}}+
\frac{\xi _3^{1/3}}{\xi _2^{1/3}}+\frac{\xi _1^{1/3}}{\xi _3^{1/3}}+
\frac{\xi _3^{1/3}}{\xi _1^{1/3}}\bigg) ={}\\
{}=\xi _1^{1/3}\xi _2^{1/3}\big(\xi _1^{1/3}+\xi _2^{1/3}\big) +
\xi _2^{1/3}\xi _3^{1/3}\big(\xi _2^{1/3}+\xi _3^{1/3}\big) +
\xi _1^{1/3}\xi _3^{1/3}\big(\xi _1^{1/3}+\xi _3^{1/3}\big)  ={}\\
{}=\big(\xi _1^{1/3}+\xi _2^{1/3}+\xi _3^{1/3}\big)
\big(\xi _1^{1/3}\xi _2^{1/3}+\xi _2^{1/3}\xi _3^{1/3}+
\xi _1^{1/3}\xi _3^{1/3}\big) -3\xi _1^{1/3}\xi _2^{1/3}\xi _3^{1/3}.
\end{multline*}
Hence and from (\ref{10}), for the respective values of
$\sqrt[3]{A},\sqrt[3]{B}$ and $\sqrt[3]{r}$ we get
\begin{align*}
\frac{\xi _1^{1/3}}{\xi _2^{1/3}}+\frac{\xi _2^{1/3}}{\xi _1^{1/3}}+
\frac{\xi _2^{1/3}}{\xi _3^{1/3}}+\frac{\xi _3^{1/3}}{\xi _2^{1/3}}&+
\frac{\xi _1^{1/3}}{\xi _3^{1/3}}+\frac{\xi _3^{1/3}}{\xi _1^{1/3}} = {}\\
&=\frac{\sqrt[3]{A}\sqrt[3]{B}}{-\sqrt[3]{r}}-3=\frac{1}{-3\sqrt[3]{r}}(A+p)-2
\end{align*}
(from the formula (3.5) in~\cite{Witula-R})
\begin{align*}
&=\frac{1}{\sqrt[3]{2}\sqrt[3]{r}}\bigg(\sqrt[3]{S+\sqrt{\tau}}+
\sqrt[3]{S-\sqrt{\tau}}\bigg) \\
&=\frac{1}{\sqrt[3]{2}}\bigg(\sqrt[3]{S_1+\sqrt{\tau _1}}+
\sqrt[3]{S_1-\sqrt{\tau _1}}\bigg) ,
\end{align*}
where
\begin{align*}
S&=rS_1,\quad \tau =r^2\tau _1, \\
S_1&=\frac{pq}{r}+\frac{6}{r^{2/3}}\big( q+p\sqrt[3]{r} +3\sqrt[3]{r^2}\big) -9, \\
\tau_1&=\bigg(\frac{pq}{r}\bigg) ^2-4\frac{q^3}{r^2}-4\frac{p^3}{r}+18\frac{pq}{r}-27= \\
&=\bigg(\frac{pq}{r}+9\bigg) ^2-\frac{4}{r^2}(q^3+p^3r+27r^2).
\end{align*}
In consequence, if $f(z)$ is the RCP polynomial
(see~\cite{Shevelev2009,WitulaFull}), then
\begin{equation}\label{11}
pr^{1/3}+3r^{2/3}+q=0,
\end{equation}
which implies
\begin{equation*}
S_1=\frac{pq}{r}-9\quad\mbox{\rm and}\quad\tau
_1=\bigg(\frac{pq}{r}+9\bigg)^2-
36\frac{pq}{r}=\bigg(\frac{pq}{r}-9\bigg) ^2,
\end{equation*}
since
\begin{equation}\label{ww12}
a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc,
\end{equation}
for $a,b,c\in\mathbb{C}$. Hence, we get the Shevelev formula
\begin{equation*}
\frac{\xi _1^{1/3}}{\xi _2^{1/3}}+\frac{\xi _2^{1/3}}{\xi _1^{1/3}}+
\frac{\xi _2^{1/3}}{\xi _3^{1/3}}+\frac{\xi _3^{1/3}}{\xi _2^{1/3}}+
\frac{\xi _1^{1/3}}{\xi _3^{1/3}}+\frac{\xi _3^{1/3}}{\xi _1^{1/3}}=
\bigg(\frac{pq}{r}-9\bigg) ^{1/3}.
\end{equation*}
However, if we assume that
\begin{equation}\label{ww13}
q^3+p^3r+27r^2=0,
\end{equation}
then we obtain
\begin{align*}
&\sqrt{ \tau_1} = \bigg|\frac{pq}{r}+9\bigg| ,\quad S_1-\frac{pq}{r}-9=
\frac{6}{r^{2/3}}\big( q+p\sqrt[3]{r}\big) , \\
&S_1+\frac{pq}{r}+9=\frac{2pq}{r}+\frac{6}{r^{2/3}}
\big( q+p\sqrt[3]{r}+3\sqrt[3]{r^2}\big) .
\end{align*}
Hence, we get the formula
\begin{multline*}
\frac{\xi _1^{1/3}}{\xi _2^{1/3}}+\frac{\xi _2^{1/3}}{\xi _1^{1/3}}+
\frac{\xi _2^{1/3}}{\xi _3^{1/3}}+\frac{\xi _3^{1/3}}{\xi _2^{1/3}}+
\frac{\xi _1^{1/3}}{\xi _3^{1/3}}+\frac{\xi _3^{1/3}}{\xi _1^{1/3}}={}\\
{}=
\sqrt[3]{\frac{3}{r^{2/3}}(q+p\sqrt[3]{r})}+\sqrt[3]{\frac{pq}{r}+
\frac{3}{r^{2/3}}(q+p\sqrt[3]{r}+3\sqrt[3]{r^2})}.
\end{multline*}
For example, let us set
\begin{equation}\label{ww14}
f(z)=z^3+3z^2-3\sqrt[3]{2}z+1.
\end{equation}
Then the condition~(\ref{ww13}) is satisfied and the roots $\xi_1$,
$\xi_2$ and $\xi_3$ of $f(z)$ are real: $\xi_1=0.56048$,
$\xi_2=0.445392$ and $\xi_3={-}4.00587$. Furthermore, the following
equality holds
\begin{multline}\label{ww15}
\frac{\xi _1^{1/3}}{\xi _2^{1/3}}+\frac{\xi _2^{1/3}}{\xi _1^{1/3}}+
\frac{\xi _2^{1/3}}{\xi _3^{1/3}}+\frac{\xi _3^{1/3}}{\xi _2^{1/3}}+
\frac{\xi _1^{1/3}}{\xi _3^{1/3}}+\frac{\xi _3^{1/3}}{\xi _1^{1/3}}=
\sqrt[3]{9(1-\sqrt[3]{2})}+\sqrt[3]{18(1-\sqrt[3]{2})}={}\\
{}=\sqrt[3]{9(1-\sqrt[3]{2})}(1+\sqrt[3]{2})={-}3.
\end{multline}
We note that condition~(\ref{ww13}), by~(\ref{ww12}), is a
condition of a type different from the condition (\ref{11}).




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\section{RCP of the second kind}

Shevelev in paper~\cite{Shevelev2007} (see also~\cite{WitulaFull})
distinguished polynomials $f\in \mathbb{R}[z]$ of the form
\begin{equation}\label{ww-gw}
f(z) = z^3 + p\, z^2 + q\, z + r,
\end{equation}
having real roots and satisfying the condition~(\ref{11}), and
called them Ramanujan cubic polynomials (shortly RCP).

Now we introduce a new family of cubic polynomials of the
form~(\ref{ww-gw}), having real roots and satisfying the
condition~(\ref{ww13}). We will call them Ramanujan cubic
polynomials of the second kind (shortly RCP2). The
polynomial~(\ref{ww14}) is an example of RCP2 which is not RCP.
On the other hand, the polynomials (see~\cite{WitulaFull}):
\begin{align*}
z^3-\frac{3}{2}\, z^2 - \frac{3}{2}\, z + 1 &= \Big( z-\frac{1}{2}\Big)\, (z+1)\, (z-2),\\
z^3+z^2 - 2\, z - 1 &= \Big( z - 2\, \cos \frac{2\, \pi}{7}\Big)\,
\Big( z - 2\, \cos \frac{4\, \pi}{7}\Big)\,
\Big( z - 2\, \cos \frac{8\, \pi}{7}\Big),
\end{align*}
belong to the set RCP$\setminus$RCP2. The polynomial 
$$z^3-3\, z+1 =
\Big( z - 2\, \cos \frac{2\, \pi}{9}\Big)\,
\Big( z - 2\, \cos \frac{4\, \pi}{9}\Big)\,
\Big( z - 2\, \cos \frac{8\, \pi}{9}\Big)
$$
belongs to the common part of families of RCP's and RCP2's
(see~\cite{WitulaFull} and Theorem~\ref{theorem2} a) written below).
The polynomial
$$
z^3 - 3\, z + \sqrt{3} =
\Big( z - 2\, \sin \frac{2\, \pi}{9}\Big)\,
\Big( z + 2\, \sin \frac{4\, \pi}{9}\Big)\,
\Big( z - 2\, \sin \frac{8\, \pi}{9}\Big)
$$
is neither RCP nor RCP2.

In the next theorem we present the basic properties of RCP2's.


\begin{theorem}\label{theorem2}
Let $f(z)\in \mathbb{R}[z]$ and be of the form~(\ref{ww-gw}).
Then the following facts hold.
\begin{list}{}{\setlength{\itemindent}{0pt}
\setlength{\labelwidth}{10pt}
\setlength{\leftmargin}{20pt}
}
\item[a)]
If $f(z)$ is either RCP or RCP2 and $p\, q\, r=0$, then $f(z)$ must
be RCP and RCP2 simultaneously.
Conversely, if $f(z)$ belongs to the  intersection of
the sets RCP and RCP2 then  $p\, q\, r \neq 0$.


\item[b)]
If $f(z)$ satisfies~(\ref{ww13}), then $f(z)$ is RCP2. In other
words, the condition~(\ref{ww13}) implies that all the roots of
$f(z)$ are real. Only in the case of $p\, q={-}9\, r$ polynomial
$f(z)$ possesses double root. In this case we have
\begin{equation}\label{ww103}
g(z):= \frac{1}{p^3}\, f(p\, z) =
z^3 + z^2 + \frac{\sqrt{5}-1}{6}\, z + \frac{1-\sqrt{5}}{54}.
\end{equation}
Moreover, if $\xi_1$ $\xi_2$ and $\xi_3=\xi_2$ are roots of $g(z)$,
then we obtain (see formula (\ref{ww91}) below):
\begin{equation}\label{ww104}
\sqrt[3]{\xi_1} + 2\, \sqrt[3]{\xi_2} =
\sqrt[3]{
{-}1 + 2\, \sqrt[3]{\frac{\sqrt{5}-1}{2}} -
6\, \sqrt[3]{
\frac{1}{3}\, \sqrt[3]{\Big(\frac{\sqrt{5}-1}{2}\Big)^2} -
\frac{1}{3}\, \sqrt[3]{\Big(\frac{\sqrt{5}-1}{2}\Big)^4}
}
}
\end{equation}
and
\begin{equation}
\sqrt[9]{\frac{\sqrt{5}-1}{2}} \,
\bigg(
1 + \sqrt[3]{\frac{\xi_1}{\xi_2}} +
\sqrt[3]{\frac{\xi_2}{\xi_1}}
\bigg) =
\sqrt[3]{\sqrt[3]{\Big(\frac{\sqrt{5}-1}{2}\Big)^2}-1}.
\end{equation}
Next, whenever $f(z)$ satisfies the condition~(\ref{11}), then $f(z)$ is
RCP if and only if $r>0$.

\item[c)]
If $f(z)$ is RCP2, then we have
\begin{equation}\label{ww15a}
r\neq 0
\quad \Longrightarrow \quad
\frac{p\, q}{r} \leqslant \frac{9}{\sqrt[3]{4}}.
\end{equation}
If $f(z)$ is RCP, then we have (see~\cite{Shevelev2007}):
\begin{equation}\label{ww15b}
r\neq 0
\quad \Longrightarrow \quad
\frac{p\, q}{r} \leqslant \frac{9}{4}.
\end{equation}

\item[d)]
If $f(z)$ is RCP, then
$$
p^2 \geqslant 12\, q,
$$
whereas, if  $f(z)$ is RCP2, then
$$
p^2 \geqslant 3\, \sqrt[3]{4}\, q.
$$

\item[e)]
Let $f(z)$ belong to family of RCP2's and let $\xi_1$, $\xi_2$,
$\xi_3$ be roots of $f(z)$. Then we have
\begin{multline}\label{ww91}
\sqrt[3]{\xi_1} +
\sqrt[3]{\xi_2} +
\sqrt[3]{\xi_3} = {}\\
{}=
\sqrt[3]{{-}p-6\, \sqrt[3]{r} - 3\,
\sqrt[3]{3\, \sqrt[3]{r}\, \big(q+p\, \sqrt[3]{r}\big)} -
3\,
\sqrt[3]{\big(p+3\, \sqrt[3]{r}\big)\, \big(q+3\, \sqrt[3]{r^2}\big)}
}.
\end{multline}
For example, for the polynomial~(\ref{ww14}) we obtain
$$
\sqrt[3]{\xi_1} +
\sqrt[3]{\xi_2} +
\sqrt[3]{\xi_3} = 0.
$$

\item[f)]
Let $f(z)$ belong to the family of RCP2's and $a, b \in \mathbb{R}$.
Suppose that $\xi_1$, $\xi_2$, $\xi_3$ are roots of $f(z)$. If $a\,
\xi_1+b$, $a\, \xi_2+b$, $a\, \xi_3+b$ are also roots of some RCP2,
then
\begin{equation}\label{ww102}
b\, \big( 9\, b^2 - 9\, a\, b\, p + a^2\, (p^2+6\, q) \big)
%\times {}\\
%{}\times
\big( 9\, b^3 - 9\, a\, b^2\, p + a^2\, b\, (p^2+6\, q) -
a^3\, (p\, q + 9\, r \big)
= 0.
\end{equation} %multline}

\item[g)]
If $a, a\, \varrho, a\, \varrho^2 \in \mathbb{R}$ are roots of some
RCP2, then
$$
2\, \sqrt[3]{2}\, \varrho = {-}\sqrt[3]{2} - 3  \pm
\sqrt{3\, \big( 3 + 2\, \sqrt[3]{2} - \sqrt[3]{4} \big)} .
$$
Moreover, for $a=2\, \sqrt[3]{2}$ we have
$$
a\, \varrho^2 =
6 - \sqrt[3]{2} + \frac{9}{\sqrt[3]{2}} \mp
\Big( 1 + \frac{3}{\sqrt[3]{2}}\Big)\,
\sqrt{3\, \Big(3 + 2\, \sqrt[3]{2} - \sqrt[3]{4}\Big)}.
$$

\item[h)]
If $f(z)$ is RCP2, then
\begin{equation}
f(z) = z^3 + \sqrt[3]{ \Big( \alpha - \frac{27}{2} \Big)\, r }\, z^2
- \sqrt[3]{ \Big( \alpha + \frac{27}{2} \Big)\, r^2 }\, z + r,
\end{equation}
for any $\alpha,r\in \mathbb{R}$.
We note that if $g(z)$ is RCP then from (18) in~\cite{WitulaFull} we have
$$
g(z) =
z^3 + \Big( \beta - \frac{3}{2} \Big) \, \varrho^{1/3}\, z^2 -
\Big( \beta + \frac{3}{2} \Big) \, \varrho^{2/3}\, z + \varrho
$$
for some $\beta,\varrho\in \mathbb{R}$.
\end{list}
\end{theorem}

\begin{proof}

\noindent
a) Both conclusions follow from (\ref{11}), (\ref{ww12}) and~(\ref{ww13}).
\bigskip

\noindent
b) Suppose that $f(z)$ satisfies~(\ref{ww13}).
Then
$$
f^{\prime}(z) =
\bigg( z + \frac{p-\sqrt{p^2-3\, q}}{3} \bigg)\,
\bigg( z + \frac{p+\sqrt{p^2-3\, q}}{3} \bigg)
$$
and
$$
f\bigg( \frac{{-}p+\sqrt{p^2-3\, q}}{3} \bigg)\,
f\bigg( \frac{{-}p-\sqrt{p^2-3\, q}}{3} \bigg)
\stackrel{(\ref{ww13})}{=}
{-}\frac{1}{27}\,
\big( p\, q + 9\, r \big)^{2},
$$
which means that all the roots of $f(z)$ are real.

Now let $p\, q ={-}9\, r$.
Then from~(\ref{ww13}) we get
$$
q^2 + \frac{3}{9}\, p^2\, q - \frac{1}{9}\, p^4 =0,
$$
which implies
$$
q=\frac{\sqrt{5}-1}{6}\, p^2
\qquad\mbox{ and }\qquad
r=\frac{1-\sqrt{5}}{54}\, p^3,
$$
and the relation~(\ref{ww103}) follows.

The equality~(\ref{ww104}) can be deduced from formula~(\ref{ww91}).
\bigskip

\noindent
c)
From~(\ref{ww13}) we get
$$
\frac{p^3\, q^3}{r^3} =
{-}\frac{p^6}{r^2} -27\, \frac{p^3}{r},
$$
which implies
\begin{equation}\label{ww100new}
\frac{9^3}{4} - \Big( \frac{p\, q}{r} \Big)^3 =
\Big( \frac{27}{2} + \frac{p^3}{r} \Big)^2 \geqslant 0,
\end{equation}
i.e.,
$$
\frac{p\, q}{r} \leqslant \frac{9}{\sqrt[3]{4}}.
$$
\bigskip

\noindent
d) From~(\ref{ww13}) we obtain
$$
27\, r^2 + p^3\, r+q^3 =0,
$$
$$
\Delta_r = p^6 - 4\cdot 27\cdot q^3 \geqslant 0,
$$
i.e,
$$
p^2 \geqslant 3\, \sqrt[3]{4}\, q.
$$
Similarly, if we have
$$
3\, \sqrt[3]{r^2} + p\, \sqrt[3]{r} + q =0
$$
and $p,q,r\in \mathbb{R}$, then
$$
\Delta_{\sqrt[3]{r}} = p^2 -12\, q \geqslant 0
\quad \Leftrightarrow \quad p^2 \geqslant 12\, q.
$$
\bigskip

\noindent
e)
We have the formula (see formula~(3.5) in~\cite{Witula-R}):
\begin{equation}\label{ww101}
\sqrt[3]{\xi_1} +\sqrt[3]{\xi_2}+\sqrt[3]{\xi_3} =
\sqrt[3]{
-p-6\, \sqrt[3]{r} - \frac{3}{\sqrt[3]{2}}\,
\Big(
\sqrt[3]{\mathcal{S}+\sqrt{\mathcal{T}}} +
\sqrt[3]{\mathcal{S}-\sqrt{\mathcal{T}}}\,
\Big)
},
\end{equation}
where
\begin{align*}
\mathcal{S} &:= p\, q +6\, q\, \sqrt[3]{r} +6\, p\, \sqrt[3]{r^2} +9\, r,\\
\mathcal{T} &:= p^2\, q^2 -4\, q^3 - 4\, p^3\, r + 18\, p\,q\, r -27\, r^2.
\end{align*}
Hence, by~(\ref{ww13}) we get
\begin{equation}\label{ww102a}
\mathcal{T} = p^2\, q^2 + 18\, p\,q\, r + 81\, r^2 =
r^2 \Big( \frac{p\, q}{r} + 9 \Big)^2,
\end{equation}
which implies
\begin{align*}
\big\{\mathcal{S} \pm \sqrt{\mathcal{T}} \big\} &=
\big\{ \mathcal{S} \pm \big(p\, q + 9\,r \big)\big\},\\
\mathcal{S} - p\, q - 9\, r &=
6\, \sqrt[3]{r}\, \big( q + p\, \sqrt[3]{r} \big),\\
\mathcal{S} + p\, q + 9\, r &=
2\, p\, q + 6\, q\, \sqrt[3]{r} +  6\, p\, \sqrt[3]{r^2} + 18\, r = \\
&=
2\, q\, \big( p + 3\, \sqrt[3]{r} \big) +
6\, \sqrt[3]{r^2}\, \big( p + 3\, \sqrt[3]{r} \big) =
2\, \big( p + 3\, \sqrt[3]{r} \big)\, \big( q + 3\, \sqrt[3]{r^2} \big)
\end{align*}
and, at last, the formula~(\ref{ww91}) follows.

In consequence, if $f(z)=z^3+3\, z^2 -3\, \sqrt[3]{2}\, z + 1$, then
$p=3$, $q={-}3\, \sqrt[3]{2}$, $r=1$ and from~(\ref{ww91}) we get
\begin{multline*}
\Big(\sqrt[3]{\xi_1} +\sqrt[3]{\xi_2}+\sqrt[3]{\xi_3} \Big)^3 =
{-}9 - 3\, \sqrt[3]{9\, (1-\sqrt[3]{2})}
- 3\, \sqrt[3]{18\, (1-\sqrt[3]{2})} = {}\\
{}=
{-}9 - 3\, \sqrt[3]{9\, (1-\sqrt[3]{2})}\,
\big( \sqrt[3]{2} +1 \big)
\stackrel{(\ref{ww15})}{=}
{-}9+9=0.
\end{multline*}
\bigskip

\noindent
f)
We have
$$
\big( x - a\, \xi_1 - b \big)\,
\big( x - a\, \xi_2 - b \big)\,
\big( x - a\, \xi_3 - b \big) =
x^3 + p_1\, x^2 + q_1 \, x + r_1,
$$
where
\begin{align*}
p_1 &= a\, p -3\, b,\\
q_1 &= a^2\, q + 3\, b^2 - 2\, a\, b\, p,\\
r_1 &= a^3\, r - a^2\, b\, q - b^3 + a\, b^2\, p.
\end{align*}
If this polynomial is also RCP2, then $q_{1}^{3} + p_{1}^{3}\, r_1 +
27\, r_{1}^{2} =0$, which (with assistance of Mathematica) implies
the equation~(\ref{ww102}).
\bigskip

\noindent
g)
Suppose that $a\neq 0$ and
$$
z^3+p\, z^2 + q\, z +  r =
(z-a)\, (z-a\, \varrho)\, (z-a\, \varrho^2).
$$
Then we have the relations
\begin{align*}
r &= {-} (a\, \varrho )^3,\\
p &= {-}a\, (1+ \varrho+\varrho^2 ),\\
q &= a^2\, (\varrho+\varrho^2+\varrho^3 ),
\end{align*}
and the condition~(\ref{ww13}) has now the form
$$
\big(\varrho+\varrho^2+\varrho^3 \big)^3 +
27\, \varrho^6 + \varrho^3\, (1+ \varrho+\varrho^2 )^3 = 0
$$
or
$$
2\, (1+ \varrho+\varrho^2 )^3 + 27\, \varrho^3 = 0.
$$
Hence
$$
\sqrt[3]{2}\, (1+ \varrho+\varrho^2 ) = {-} 3\, \varrho,
$$
$$
\varrho^2 + \Big( 1 + \frac{3}{\sqrt[3]{2}} \Big)\, \varrho + 1 = 0,
$$
which implies
$$
2\, \sqrt[3]{2}\, \varrho = {-} \sqrt[3]{2} - 3 \pm
\sqrt{3\, \big( 3 + 2\, \sqrt[3]{2} - \sqrt[3]{4}  \big)}.
$$
\bigskip

\noindent
h) Let us set
\begin{equation}\label{ww101new}
\alpha:= \frac{27}{2} + \frac{p^3}{r}.
\end{equation}
Then from~(\ref{ww100new}) we generate the relation
$$
\frac{9^3}{4} -
\Big( \alpha - \frac{27}{2} \Big)\, \frac{q^3}{r^2} = \alpha^2,
$$
i.e.,
$$
\frac{9^3}{4} - \alpha^2 =
\Big( \alpha - \frac{27}{2} \Big)\, \frac{q^3}{r^2},
$$
$$
q^3 =
{-}\Big( \alpha + \frac{27}{2} \Big)\, r^2.
$$
From~(\ref{ww101new}) we obtain
$$
p^3 = \Big( \alpha - \frac{27}{2} \Big)\, r.
$$

\end{proof}

The following theorem, proved by Shevelev for
RCP's~\cite{Shevelev2007}, holds also for RCP2's.


\begin{theorem}
If for two RCP2's of the form
$$
y^3 + p_1\, y^2 + q_1\, y + r_1,\qquad
z^3 + p_2\, z^2 + q_2\, z + r_2
$$
the following condition holds ($r_1\, r_2 \neq 0$):
$$
\frac{p_1\, q_1}{r_1} = \frac{p_2\, q_2}{r_2},
$$
then for their roots $y_1$, $y_2$ $y_3$ and $z_1$, $z_2$, $z_3$,
respectively, the sequence of numbers
$$
\frac{y_1}{y_2},
\frac{y_2}{y_1},
\frac{y_1}{y_3},
\frac{y_3}{y_1},
\frac{y_2}{y_3},
\frac{y_3}{y_2},
$$
is a~permutation of the sequence
$$
\frac{z_1}{z_2},
\frac{z_2}{z_1},
\frac{z_1}{z_3},
\frac{z_3}{z_1},
\frac{z_2}{z_3},
\frac{z_3}{z_2}.
$$
\end{theorem}

\begin{proof}
The proof runs like Shevelev's proof of Theorem~5
in~\cite{Shevelev2007}. Only one change is needed, for the case of
RCP2 in formula (38) we have
$$
\frac{p^3\, r + q^3}{r^2}
\stackrel{(\ref{ww13})}{=}
{-}27.
$$
\end{proof}

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%


\section{Acknowledgments}

I wish to express my gratitude to the referee
for several helpful comments and suggestions
concerning the first version of the paper.


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%%%%%%%%%%%%%%%%%%%%%%%%%%              The Bibliography              %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{thebibliography}{99}

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\bibitem{Witula-R}
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\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: Primary
11C08; Secondary 11B83, 33B10.

\noindent \emph{Keywords:} Ramanujan cubic polynomial.

\bigskip
\hrule
\bigskip

\vspace*{+.1in} \noindent Received June 17 2010;  revised version
received July 1 2010.
Published in {\it Journal of Integer Sequences}, July 9 2010.

\bigskip
\hrule
\bigskip

\noindent Return to \htmladdnormallink{Journal of Integer Sequences
home page}{http://www.math.uwaterloo.ca/JIS/}. \vskip .1in


\end{document}
