If the complex roots of $x^3-x-2=0$ are $r\pm si$, and $As^6 +Bs^4 + Cs^2 =26$ for integers $A$, $B$, $C$, find $A+B+C$

Question

The question:

In the cubic $x^3-x-2=0$, there is one real root and two complex roots of the form $r\pm si$, with $r$ and $s$ real. If there exists integers $A,B,$ and $C$ such that $As^6 +Bs^4 + Cs^2 =26$, find A+B+C.

What I've done: I expanded $(x-a)(x-r-si)(x-r+si)$ where $a$ is the real root, and matched coefficients with $x^3-x-2$ to get relationships such as $-2x^2r-ax^2=0$, upon which $a=-2r$. Another (I think) important relationship I got was $r^2+s^2=\dfrac{2}{a}$. Using these, i substituted $r= \dfrac{-a}{2}$ into $r^2+s^2 = \dfrac{2}{a}$ to get that $s^2 = \dfrac{2}{a} - \dfrac{a^2}{4}$. However, substituting this into $As^6 + Bs^4 + Cs^2 = 26$, I have not been able to make progress.

Answer 1

You could also solve your problem in the following way.

Since $\,r+is\;(\text{where }s\neq0)\,$ is a complex root of the equation $\,x^3-x+2=0\,,\;$ it follows that

$(r+is)^3-(r+is)+2=0\;\;,$

$r^3+3ir^2s-3rs^2-is^3-r-is-2=0\;\;,$

$r^3-3rs^2-r-2+is\!\left(3r^2-s^2-1\right)=0\;\;,$

$\begin{cases}r^3-3rs^2-r-2=0\\[3pt]3r^2-s^2-1=0\end{cases}$

$\begin{cases}r\left(r^2-3s^2-1\right)=2\\[3pt]r^2=\dfrac{s^2+1}3\end{cases}$

$\begin{cases}r\left(\dfrac{s^2+1}3-3s^2-1\right)=2\quad\implies r=\dfrac{-3}{4s^2+1}\\[3pt]s^2+1=3r^2\quad\color{blue}{(*)}\end{cases}$

$r^2=\dfrac9{16s^4+8s^2+1}\quad\color{blue}{(**)}$

From $\,(*)\,$ and $\,(**)\,$ it follows that

$\left(s^2+1\right)\left(16s^4+8s^2+1\right)=27\;\;,$

$16s^6+24s^4+9s^2=26\;.$

Consequently ,

$A=16\,,\;B=24\,,\;C=9\;,\;$ hence ,

$A+B+C=49\,.$

Answer 2

We use Vieta's formulas to get an expression for $s$ in terms of the real root $a$, where $a^3-a-2=0$.

Then

$a+(r+si)+(r-si)=0\implies r=-a/2$

$a(r+si)(r-si)=2\implies a(r^2+s^2)=2\implies r^2+s^2=2/a$

Thus

$s^2=(2/a)-(a/2)^2=(8-a^3)/4a$

But $a^3=a+2$ since $a$ is the real root of the cubic equation. So we simply have

$s^2=(6-a)/4a; a=6/(4s^2+1)$

Then $a^3-a-2=0\implies\left(\dfrac6{4s^2+1}\right)^3-\dfrac6{4s^2+1}-2=0$

Clearing fractions and expanding the powers of $s^2+1$ leads to

$128s^6+192s^4+72s^2-208=8(16s^6+24s^4+9s^2-26)=0$

So $\color{blue}{16}s^6+\color{blue}{24}s^4+\color{blue}{9}s^2=26$, whence $A+B+C=49$.