If $\alpha$ be a root of the equation $x^3+3x^2-6x+1=0$,prove that the other roots are $\frac 1{1-\alpha},\frac {\alpha - 1}\alpha$.

Question

If $\alpha$ be a root of the equation $x^3+3x^2-6x+1=0$, then find the other roots. $\bf{Try}: $ I tried by using relation between roots and coefficients. Let $\beta, \gamma$ be the other two roots. Then
$\sum \alpha =-3$
$ \sum \alpha\beta=-6$

$\alpha\beta\gamma=-1$ But unable to find $\beta, \gamma $ in the desired form from these three equations.
Hope somebody will teach me some cleaver ways to tackle this type of problems. Thanks in advance. $\bf{Note} $ : Edited after Aqua's comment.

Answer 1

What you can do is try transforming the whole polynomial. Try creating a polynomial whose roots are $\frac{\alpha - 1}{\alpha}, \frac{\beta - 1}{\beta}, \frac{\gamma - 1}{\gamma}$. We can form a polynomial with roots $\alpha^{-1}, \beta^{-1}, \gamma^{-1}$ by simply reversing the order of the coefficients: $$x^3 - 6x^2 + 3x + 1.$$ Then, we just need to subtract $1$ from each root, which we do by replacing $x$ with $x + 1$: $$(x + 1)^3 - 6(x + 1)^2 + 3(x + 1) + 1 = x^3 - 3x^2 - 6x - 1.$$ The roots of this polynomial are the roots of the original polynomial mapped under $\lambda \mapsto \lambda^{-1} - 1$. To get the polynomial corresponding to the map $\lambda \mapsto 1 - \lambda^{-1} = \frac{\lambda - 1}{\lambda}$, we take the above polynomial and substitute $x$ with $-x$. This produces $$(-x)^3 - 3(-x)^2 - 6(-x) - 1 = -x^3 - 3x^2 + 6x - 1,$$ which equals zero if and only if the original polynomial equals $0$.

Thus, the roots of the original polynomial are stable under the map $\lambda \mapsto \frac{\lambda - 1}{\lambda}$. Thus, if $\alpha$ is a root, then so is $\frac{\alpha - 1}{\alpha}$, and $$\frac{\frac{\alpha - 1}{\alpha} - 1}{\frac{\alpha - 1}{\alpha}} = \frac{-\alpha^{-1}}{\frac{\alpha - 1}{\alpha}} = \frac{-1}{\alpha - 1} = \frac{1}{1 - \alpha}.$$ The only thing remaining to check is that these three roots are distinct. If we have some root $\alpha$ such that $\alpha = \frac{\alpha - 1}{\alpha}$, then $\alpha^2 - \alpha + 1 = 0$. Testing the roots of this quadratic manually (or simply attempting to divide one polynomial from the other) shows that this polynomial has no such root.

Answer 2

Let

$$p(x)=x^3+3x^2-6x+1=0$$

I propose two solutions:

1. A first one using a Galois theory property.

Observe first that function $f$ defined by $f(x)=\dfrac{1}{1-x}$ generates the following 3-elements group of homographies:

$$f(x), \ \ \ \ f^{(2)}(x):=f(f(x))=\dfrac{x-1}{x}=f^{-1}(x), \ \ \ \ f^{(3)}(x):=f(f(f(x)))=x, $$

itself a subgroup of the cross-ratio group (isomorphic to the 6-element symmetric group $S_3$) described in my answer here.

The discriminant of polynomial $p$ is

$$729=27^2$$

In such a case (see this document), the Galois group associated to polynomial $p$ is the 3-elements alternate group $A_3$. Therefore, as there is a unique (cyclic) structure for a 3-elements group, these two groups must coincide.

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2. A computational proof:

Let

$$u:=a; \ \ \ \ v:=\dfrac{1}{1-a}; \ \ \ \ w:=\dfrac{a-1}{a} \tag{0}$$

$$u+v+w+3=\dfrac{a^3 + 3a^2 - 6a + 1}{a(a - 1)} \tag{1}$$

which is $0$ because $a$ is a root of polynomial $p$.

$$uv+uw+vw+6=\dfrac{a^3 + 3a^2 - 6a + 1}{a(a - 1)} \tag{2}$$

equal to $0$ for the same reason.

$$uvw+1=0 \tag{3}$$

(1),(2) and (3) being equivalent to Vieta's relationships

$$u+v+w=-3, \ \ uv+vw+wu=-6, \ \ uvw=-1$$

attached to polynomial $p$, and by unicity of this correspondence, we are able to conclude that $u,v,w$ given by (0) are the roots of $p$.

Answer 3

This really seems to be simply intended as a "verification" type of problem, in which you are just asked to show that two roots of a particular form satisfy the cubic equation. [You can accomplish that by just inserting the other two given roots into the polynomial.] The concern that you expressed in the comments that the poser of the problem might ask for the relation itself among the roots is unnecessary, since there is no straightforward way to "extract" that relation from the polynomial itself.

Consider that working from, say, the Viete relations, we only find $$ \alpha + r_1 + r_2 = -3 \ \ , \ \ \alpha r_1 + \alpha r_2 + r_1 r_2 = 6 \ \ , \ \ \alpha r_1 r_2 = -1 \ \ \Rightarrow \ \ \alpha · (-3 - \alpha) \ + \ \left( -\frac{1}{\alpha} \right) \ = \ -6 \ \ , $$

which just gives you back $ \ -\alpha^3 - 3 \alpha^2 + 6 \alpha - 1 \ = \ 0 \ \ . $ Even if you suspected that the three roots had the relation $ \ \alpha \ , \ \frac{1}{f(x)} \ , \ -\frac{f(x)}{\alpha} \ , $ the coefficients alone do not provide enough information to divine what $ \ f(x) \ $ might be.

The sort of thing one could say more about is if the form of the roots is $ \ \alpha \ , \ \frac{1}{p \alpha + q} \ , \ -\frac{p \alpha + q}{\alpha} \ , $ maintaining the product of the roots as $ \ -1 \ , $ then for the polynomial $ \ x^3 + b x^2 + c x + 1 \ \ , $ we have (in the manner of Jean Marie):

$$ \alpha + \left(\frac{1}{p\alpha + q} \right) + \left( -\frac{p\alpha + q}{\alpha} \right) + b \ \ = \ \ \frac{p\alpha^3 \ + \ (bp - p^2 + q)·\alpha^2 \ + \ (bq - 2pq + 1)·\alpha \ - \ q^2}{\alpha · (p\alpha + q)} \ \ = \ \ 0 \ \ , $$

$$ \ \ \left(\frac{\alpha}{p\alpha + q} \right) - (p\alpha + q) + \left(-\frac{1}{ \alpha } \right) - c \ \ = \ \ \frac{-p^2\alpha^3 \ - \ (cp + 2pq - 1)·\alpha^2 \ - \ (p + cq + q^2)·\alpha \ - \ q}{\alpha · (p\alpha + q)} \ \ = \ \ 0 \ \ $$

With $ \ p = -1 \ $ and $ \ q = 1 \ , $ for the roots in this problem, the numerator in the first equation gives us $ \ -\alpha^3 - b·\alpha^2 + (b + 3)·\alpha - 1 \ = \ 0 \ $ and the numerator in the second yields $ \ -\alpha^3 + (c+3)·\alpha^2 - c·\alpha - 1 \ = \ 0 \ \ . $ The relation between the roots is thus associated with a family of cubic polynomials $ \ x^3 \ + \ bx^2 \ - \ (b+3)x \ + \ 1 \ $ or $ \ x^3 \ - \ (c+3)x^2 \ + \ cx \ + \ 1 \ \ ; $ if we choose $ \ b = 3 \ \ \text{or} \ \ c = -6 \ \ , $ we then obtain the specified polynomial $ \ x^3 \ + \ 3x^2 \ - \ 6x \ + \ 1 \ \ . $

$$ \ $$

We can use some results from the theory of equations to see what might be said about the roots in question. Jean Marie shows the cubic discriminant to be $ \ 729 \ $ (I also find this value), so this positive value tells us that the polynomial has three distinct real roots (Cardano's casus irreducibilis). The Rule of Signs indicates that two of these are positive and one negative; since both the leading coefficient and the constant term equal $ \ 1 \ , $ the only rational root candidates are $ \ \pm 1 \ , $ plainly neither of which are zeroes.

The location of the polynomial's local extrema are found from $$ 3x^2 \ + \ 6x \ - \ 6 \ = \ 0 \ \ \Rightarrow \ \ x \ = \ -1 - \sqrt{3} \ \approx -2.732 \ \ , \ \ \sqrt{3} - 1 \ \approx 0.732 \ \ . $$ So the zeroes are found in the intervals $ \ x < -1 - \sqrt{3} \ , \ 0 < x < \sqrt{3} - 1 \ , \ x > \sqrt{3} - 1 \ \ . $

We can attempt to determine which of the described zeroes lies in each interval. If we assign

$$ \alpha < -1 - \sqrt{3} \ \ \Rightarrow \ \ 0 \ < \ \frac{1}{1 - \alpha} \ < \ \frac{1}{2 + \sqrt{3}} \ \approx \ 0.368 \ , \ 0 \ < \ 1 - \frac{1}{\alpha} \ < \ \frac{2 + \sqrt{3}}{1 + \sqrt{3}} \ \approx \ 1.366 \ \ . $$

This "labeling" is consistent with the intervals we've determined. However, for the alternative choices, we find

$$ 0 \ < \ \alpha \ < \ \sqrt{3} - 1 \ \ \Rightarrow \ \ 0 \ < \frac{1}{1 - \alpha} \ < \ \frac{1}{2 - \sqrt{3}} \ \approx \ 3.732 \ , \ \ 1 - \frac{1}{\alpha} \ < \ \frac{2 - \sqrt{3}}{1 - \sqrt{3}} \ \approx -0.366 $$ and $$ \alpha \ > \ \sqrt{3} - 1 \ \ \Rightarrow \ \ \frac{1}{1 - \alpha} \ > \ \frac{1}{2 - \sqrt{3}} \ , \ \ 1 - \frac{1}{\alpha} \ > \ \frac{2 - \sqrt{3}}{1 - \sqrt{3}} \ \ , $$

for which the second set of inequalities also is consistent with the required intervals, and the third set is mostly consistent. This suggests the symmetry of exchange of the zeroes that is described more precisely by Jean Marie and Theo Bendit. We can observe this symmetry by calculating with the (approximate) irrational zeroes ( $ -4.4115 \ , \ 0.1848 \ , \ 1.2267 $ ) found using a resource such as WolframAlpha.

Answer 4

The cubic is equivalent to $(x+1)^3=9x$. If $(\alpha+1)^3=9\alpha$ then$$(2-\alpha)^3=9(1-\alpha)^2,\,(2\alpha-1)^3=9\alpha^2(\alpha-1)$$because$$(2-\alpha)^3-9(1-\alpha)^2=(2\alpha-1)^3-9\alpha^2(\alpha-1)=-\alpha^3-3\alpha^2+6\alpha-1.$$