1

Problem: Consider the polynomial $f(x) = x^3 + x^2 - 4x + 1$

a) Show that if $r$ is a root of $f$, then $r^2 + r -3$ is also a root of $f$.

b) Let, $\alpha, \beta, \gamma$ be the three roots of $f$. Determine all possible values of $\frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\alpha}$

For the first part of this problem, I just plugged in the given expression and factorised, and it gave me $f(r)$ in the factorised expression so that was done, but it was extremely lengthy.

For the second part, I actually don't have much of a clue (the best I could do was obtain $\frac{\alpha^2 \beta + \beta^2 \gamma + \gamma^2 \alpha}{\alpha \beta \gamma}$, and hints regarding part (b) is my main problem.

Shivansh Jaiswal
  • 86
  • 3
    This question might be relevant. In short: If $u=\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$ and $v=\alpha^2\gamma+\beta^2\alpha+\gamma^2\beta,$ then $u+v$ and $uv$ are symmetric in $\alpha,$ $\beta$ and $\gamma.$ Therefore, you can derive $u+v$ and $uv$ from the coefficients of the given polynomial. You can then get $u$ and $v$ solving a quadratic equation. As the order of the roots is not specified, $u$ and $v$ are both valid solutions. $u$ becomes $v$ and $v$ becomes $u$ if you swap two of the roots. – Reinhard Meier Aug 14 '23 at 12:48
  • For the first part, perhaps they want you to notice that if $r$ is a root, then $r^2+r-3=1-\frac{1}{r}$; this is an easier substitution. For the second part, it looks like you're nearly there - dividing through your expression is very similar to what's needed. I assume you're using Vieta for that? – Chris Lewis Aug 14 '23 at 12:51
  • see https://math.stackexchange.com/questions/2022216/on-the-trigonometric-roots-of-a-cubic as your cubic has square discriminant – Will Jagy Oct 24 '23 at 04:46

3 Answers3

3

Here is another way, using (a) to solve (b):

a) We have $x^2+x-4+\dfrac1x=0 \implies x^2+x-3=1-\dfrac1x$. Hence $$f(r^2+r-3)=f\left(1-\frac1r\right) = \dfrac{(r-1)^3+r(r-1)^2-4r^2(r-1)+r^3}{r^3}\\=-\frac{f(r)}{r^3}=0.$$

b) Given (a), the roots must be of form $p=r^2+r-3, q=p^2+p-3, r= q^2+q-3$. Then, $\frac{p}q+\frac{q}r+\frac{r}p=$ $$\frac{p^2r+q^2p+r^2q}{pqr}=-((q-p+3)r+(r-q+3)p+(p-r+3)q)\\=-3\sum_{cyc} p=3$$

Alternately, changing the order non-cyclically, $\frac{q}p+\frac{p}r+\frac{r}q=$ $$\frac{q^2r+p^2q+r^2p}{pqr}=-((r-q+3)r+(q-p+3)q+(p-r+3)p)\\=-\sum_{cyc} p^2+\sum_{cyc} pq-3\sum_{cyc} p = -10$$

Macavity
  • 46,381
2

Part (b)

Let $u = \alpha^2 \beta + \beta^2 \gamma + \gamma^2 \alpha$, and $v = \alpha \beta^2 + \beta \gamma^2 + \gamma \alpha^2$. Define $$ e_1 = \frac{-b}{a} = \sum_{cyc}{\alpha} $$ $$ e_2 = \frac{c}{a} = \sum_{cyc}{\alpha \beta} $$ $$ e_3 = \frac{-d}{a} = \alpha \beta \gamma $$

In this case, the values are $e_1 = -1, e_2 = -4, e_3 = - 1$.

Consider the following expression,

$$ \frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\alpha} = \frac{\alpha \beta^2 + \beta \gamma^2 + \gamma \alpha^2}{\alpha \beta \gamma} = \frac{v}{e_3} = -v $$

By using the following expressions for $u + v$ and $uv$

$$ u + v = e_1 e_2 - 3 e_3 = 7 $$ $$ uv = e_1 ^3 e_3 - 6e_1 e_2 e_3 + e_2^3 + 9e_3 ^3 = -30$$

Thus we have the equation,

$$ v^2 - 7v - 30 = v^2 - 10v + 3v - 30 = (v-10)(v+3) $$

Thus, $v = +10, -3$, hence the result

$$ \frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\alpha} = -v = -10, +3$$

Shivansh Jaiswal
  • 86
0

HINT.- Another solution is as follows: one has the two possible values to calculate $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$ and $\alpha^2\gamma+\beta^2\alpha+\gamma^2\beta$ where $\alpha=r,\space \beta=r^2+r-3,\space \gamma=-r^2-2r+2$. For this we have $$\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha=P(r)$$ where $P(r)$ is a polynomial in $r$ of $6$-degree. But we have that all power $p^k$ with $k\ge3$ is a linear combination of $1,r,r^2$, namely one has $$r^3=-r^2+4r-1\\r^4=5r^2-5r+1\\ r^5=-10r^2+21r-5\\r^6=31r^2-45r+10$$ Consequently $$\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha=Ar^2+Br+C$$ The corresponding value will be $C$ and the coefficients $A$ and $B$ should be zero. Certainly for $\alpha^2\gamma+\beta^2\alpha+\gamma^2\beta$ it will in general be another value $C'$ but $A'=B'=0$

Piquito
  • 29,594