Rotating the roots of a cubic

Let $f(x) = x^3 + ax^2 + bx + c , f(0) = c = 2 , \\ f(1) = 1 + a + b + c = 3 + a + b = 9 \Rightarrow a + b = 6.$ Now , Let $g(x) = px + q .$ $g(\alpha_{1}) = p\alpha_{1} + q = \alpha_{2}.....(i), \\ g(\alpha_{2}) = p\alpha_{2} + q = \alpha_{3}.......(ii) , and \\ g(\alpha_{3}) = p\alpha_{3} + q = \alpha_{1}......(iii).$ Using $(i)$ and $(ii)$ , we get p = $\frac{\alpha_{2} - \alpha_{3}}{\alpha_{1} - \alpha_{2}}$ , Using $(ii)$ and $(iii)$ , we get p = $\frac{\alpha_{3} - \alpha_{1}}{\alpha_{2} - \alpha_{3}}$ . Equating the two , $(\alpha_{1})^2 + (\alpha_{2})^2 + (\alpha_{3})^2 = \alpha_{1}\alpha_{2} + \alpha_{2}\alpha_{3} + \alpha_{3}\alpha_{1} \\ \Rightarrow (\alpha_{1} + \alpha_{2} + \alpha_{3})^2 = 3( \alpha_{1}\alpha_{2} + \alpha_{2}\alpha_{3} + \alpha_{3}\alpha_{1}) \Rightarrow a^2 = 3b$ , also, $b = 6 - a \Rightarrow a^2 = 3( 6 - a) \Rightarrow a^2 + 3a - 18 = 0 \Rightarrow \fbox{ a = 3}$ or $\fbox{ a = -6}$ Correspondingly , $\fbox{ b = 3 }$ , or $\fbox{b = 12}$ Put these values to get , $f(x) = x^3 + 3x^2 + 3x + 2$ or $f(x) = x^3 - 6x^2 + 12x +2 \Rightarrow f(2) = 28$ or $f(2) = 10.$ Hence, sum of possible values = 28 + 10 = $\fbox{38}$

Suppose that $g(x)=ax+b$ . Then $\begin{array}{rcl} \alpha_2 & = & a\alpha_1 + b \\ \alpha_3 & = & a\alpha_2 + b \; = \; a^2\alpha_1 + (a+1)b \\ \alpha_1 & = & a\alpha_3 + b \; = \; a^3\alpha_1 + (a^2+a+1)b \end{array}$ and so $(a^2+a+1)\big[(a-1)\alpha_1 + b\big] \; = \; (a^2 + a + 1)(\alpha_2-\alpha_1) \; = \; 0$ Since we could replace $g$ by $g^{-1}$ , we can assume without loss of generality that $a = \omega$ , the principal primitive cube root of unity.

Introducing $c = \tfrac{b}{1-\omega}$ , we note that $f(c)=c$ . We can write $\alpha_1 \; = \; c + d \qquad \alpha_2 \; = \; c + \omega d \qquad \alpha_3 \ = \; x + \omega^2d$ where $d \neq 0$ . Thus $f(x) \,=\, (x-c)^3 - d^3$ . Since $-c^3 - d^3 \; = \; f(0) \; = \; 2 \qquad \qquad (1-c)^3 - d^3 \; = \; f(1) \; = \; 9$ we deduce that $c^3 + (1-c)^3 \,=\, 7$ , or $c^2 - c - 2 \,=\, 0$ , so that $c = 2,-1$ .

• If $c=2$ then $d^3 = -10$ , so that $f(x) = (x-2)^3 + 10$ . Thus $f(2) = 10$ .

• If $c=-1$ then $d^3 = -1$ , so that $f(x) = (x+1)^3 + 1$ . Thus $f(2) = 28$ .

Adding these answers together, we deduce that $N = 38$ .

From the first condition, we deduce that $f(x)= (x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$

Since $f(0)=2$ , we have $(-\alpha_1)(-\alpha_2)(-\alpha_3)=2$ , or $\alpha_1 \alpha_2 \alpha_3 =-2$

Since $f(1)=9$ , we have $(1-\alpha_1)(1-\alpha_2)(1-\alpha_3)=9$ . Expanding the brackets gives us $1-(\alpha_1+\alpha_2+\alpha_3)+(\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_1\alpha_3)-\alpha_1 \alpha_2 \alpha_3=9$ , and since we know that $\alpha_1 \alpha_2 \alpha_3=-2$ , this simplifies to $-(\alpha_1+\alpha_2+\alpha_3)+(\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_1\alpha_3) = 6 (*)$

Now we focus on the second condition. Let $g(x)=mx+b$ , then we have $m\alpha_1+b=\alpha_2$ , $m\alpha_2+b=\alpha_3$ , and $m\alpha_3+b=\alpha_1$ . Subtracting the first equation from the second equation gives us $m(\alpha_2-\alpha_1)=\alpha_3-\alpha_2$ and subtracting the second equation from the third equation gives us $m(\alpha_3-\alpha_2)=\alpha_1-\alpha_3$ .

Since $\alpha_1,\alpha_2,\alpha_3$ are distinct, we thus have $m=\frac{\alpha_3-\alpha_2}{\alpha_2-\alpha_1}=\frac{\alpha_1-\alpha_3}{\alpha_3-\alpha_2}$

Cross-multiplying, we have $(\alpha_3-\alpha_2)^2=(\alpha_2-\alpha_1)(\alpha_1-\alpha_3)$ , and after simplification, this gives us $\alpha_1^2+\alpha_2^2+\alpha_3^2=\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_3\alpha_1$ , which is equivalent to $(\alpha_1+\alpha_2+\alpha_3)^2=3(\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_3\alpha_1)$

If we let $A=\alpha_1+\alpha_2+\alpha_3$ and $B=\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_3\alpha_1$ , we would have $A^2=3B$ . Furthermore from equation (*), $B=A+6$ . Thus, $A^2=3(A+6)$ , implying $A^2-3A-18=0$ , which means $A=6$ or $A=-3$

Therefore we have two cases: $A=6,B=12$ or $A=-3,B=3$ . Using Vieta's theorem: $f(x)=x^3-(\alpha_1+\alpha_2+\alpha_3)x^2+(\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_3\alpha_1)x-\alpha_1\alpha_2\alpha_3$ , we arrive at two possible functions:

$f(x)=x^3-6x^2+12x+2$ or $f(x)=x^3+3x^2+3x+2$

In the first case, $f(2)=10$ and in the second case, $f(2) = 28$ . Thus the sum of the possible values of $f(2)$ is $10+28=\fbox{38}$

A polynomial $f$ has the form $f(z) = z^3 - az^2 + b z - c,$ where $a = \alpha_1 + \alpha_2 + \alpha_3,\, b = \alpha_1\alpha_2+\alpha_2\alpha_3 + \alpha_3\alpha_1,$ and $c=\alpha_1\alpha_2\alpha_3.$ From conditions 3 and 4, we have $c=-2\quad \text{and}\quad b-a= 6.\quad\quad(*)$ Let $g(z) = \alpha z + \beta.$ Condition 2 implies $g^3(\alpha_i)=g(g(g(\alpha_i))) = \alpha_i\quad \text{for}\,\, \, i=1,2,3.$ Together with the fact that $g^3(z) = \alpha^3 z + \beta(\alpha^2+\alpha+1)$ we have $\alpha^3 \alpha_i + \beta(\alpha^2+\alpha+1) = \alpha_i \quad \text{for} \,\,\, i=1,2,3.$ Subtracting the above equations with $i =1$ and $i=2$ we have $\alpha^3(\alpha_1 - \alpha_2) = \alpha_1-\alpha_2.$ By the assumption that $\alpha_1 \neq \alpha_2$ we have $\alpha^3 = 1.$ This implies $\alpha$ is a cube root of unity. If $\alpha =1,$ then one of the above equations yields $\alpha_1 + 3\beta = \alpha_1,$ so that $\beta = 0$ and we have $g(z) = \alpha z + \beta = z$ implying $\alpha_2 = g(\alpha_1) = \alpha_1,$ a contradiction. Hence $\alpha_1 \neq 1.$ This implies $\alpha$ is a primitive cube root of unity. Let's denote it by $\omega.$ Hence $g(z) = \omega z + \beta$ or $\beta = g(z) -\omega z.$ Substituting $z=\alpha_i$ for $i=1,2,3$ into this equation, we have $\alpha_2-\omega \alpha_1 = \alpha_3-\omega \alpha_2 = \alpha_1 - \omega \alpha_3.$ The first pair of the equations yields $(1+\omega)\alpha_2 = \omega \alpha_1+\alpha_3$ or $\omega \alpha_1 +\omega^2 \alpha_2 + \alpha_3 = 0,$ since $\omega^2+\omega+1 = 0.$ We have similar relations with the other two pairs of the equations. We can translate these relations into a matrix equation as follows: $AX = \mathbf{0},$ where $A=\begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \\ \alpha_2 & \alpha_3 & \alpha_1\\ \alpha_3 & \alpha_1 & \alpha_2\end{pmatrix}$ and $X= \begin{pmatrix} 1 \\ \omega \\ \omega^2 \end{pmatrix}.$ Since $X\neq\mathbf{0},$ this implies $\det A = 0.$ (Otherwise $A$ is invertible yielding $X = A^{-1}\mathbf{0} = \mathbf{0},$ a contradiction.) Adding the second and third rows to the first row, and factoring, we have $\det A = (\alpha_1+\alpha_2+\alpha_3)\begin{vmatrix} 1 & 1 & 1 \\ \alpha_2 & \alpha_3 & \alpha_1 \\ \alpha_3 & \alpha_1 & \alpha_2 \end{vmatrix}.$ Expanding along the first row, we obtain $\det A = \sum_{\mathrm{cyc}}\alpha_1\Big(\sum_{\mathrm{cyc}}\alpha_1\alpha_2 - \sum_{\mathrm{cyc}}\alpha_1^2\Big).$ Imposing $\det A = 0$ yields either terms of the product to be $0$ . If $\sum_{\mathrm{cyc}}\alpha_1 = 0,$ then subtracting from the relation $\omega\alpha_1 +\omega^2\alpha_2 +\alpha_3 = 0$ yields $\alpha_1 + (\omega+1)\alpha_2 = 0$ or $\alpha_1 = \omega^2 \alpha_2$ or $\omega \alpha_1 = \alpha_2,$ where the last two equations follow from the facts that $\omega^2 + \omega + 1 = 0$ and $\omega^2 = \frac{1}{\omega}.$ Similarly, we obtain $\omega \alpha_2 = \alpha_3$ and $\omega \alpha_3 = \alpha_1.$ Multiplying these three equations by $\alpha_3,\alpha_1,$ and $\alpha_2,$ respectively and combining, we have $\omega\sum_{\mathrm{cyc}}\alpha_1\alpha_2 = \sum_{\mathrm{cyc}}\alpha_1\alpha_2.$ Since $\omega \neq 0,$ we have $\sum_{\mathrm{cyc}}\alpha_1\alpha_2 =0.$ In conclusion, we have $a=\sum_{\mathrm{cyc}}\alpha_1 = 0 \quad \text{and} \quad b = \sum_{\mathrm{cyc}}\alpha_1\alpha_2 = 0,$ contradicting the equation $(*)$ above.

Therefore $\sum_{\mathrm{cyc}}\alpha_1\alpha_2 - \sum_{\mathrm{cyc}}\alpha_1^2 = 0.$ In terms of $a$ and $b,$ this is $3b-a^2 = 0.$ Combining with the equation $(*)$ we have $(a,b) = (6,12),\,(-3,3).$ With $c=-2$ , this yields two possible polynomials $f,$ namely $f(z) =z^3-6z^2+12z+2 \quad \text{and} \quad f(z) =z^3+3z^2+3z+2.$ This gives two possible values of $f(2)$ . They are $f(2) = 2^3-6(2^2)+12(2)+2 = 10$ and $f(2) = 2^3+3(2^2)+3(2)+2 = 28.$ Hence the sum of all possible values of $f(2)$ is $10+28 = \boxed{38}.$

This is not the way I came to the solution, but these steps are a bit more motivated.

Using 1. condition ( f(x) is monic) , we can write:

f(x) = (x - α1)(x - α2)(x - α3).

From 3. and 4. we have:

-α1 α2 α3 = 2 ...(#) and

(1 - α1)(1 - α2)(1 - α3) = 9 which can be written as:

1 - (α1 + α2 + α3) + (α1 α2 + α1 α3 + α2 α3) - α1 α2 α3 = 9,

using (#) we get

-(α1 + α2 + α3) + (α1 α2 + α1 α3 + α2 α3) = 6 (##)

We are asked for f(2):

f(2) = (2 - α1)(2 - α2)(2 - α3)

f(2) = 8 - 4(α1 + α2 + α3) + 2(α1 α2 + α1 α3 + α2 α3) - α1 α2 α3,

and using (#)

f(2) = 10 - 4(α1 + α2 + α3) + 2(α1 α2 + α1 α3 + α2 α3)

Here we realize that we only need to find (α1 + α2 + α3) and

(α1 α2 + α1 α3 + α2 α3) if possible.

Now we use condition 2. writing down the g(x):

g(x) = Ax + B, so

A α1 + B = α2 ...(1)

A α2 + B = α3 ...(2)

A α3 + B = α1 ...(3)

We do not need A and B, so we try to eliminate them, first A then B.

Multiplying (1) with α2 and (2) with α1 and taking difference (and extracting B):

B (α2 - α1) = α2^2 - α3 α1

Doing the same with (2) and (3) and also with (3) and (1), always eliminating A we get:

B (α3 - α2) = α3^2 - α1 α2

B (α1 - α3) = α1^2 - α2 α3

Summing these 3 equations and rearranging we get:

α1^2 + α2^2 + α3^2 = α1 α2 + α2 α3 + α3 α1 ... (###)

Now we can write a known relation:

(α1 + α2 + α3)^2 = α1^2 + α2^2 + α3^2 + 2(α1 α2 + α2 α3 + α3 α1),

using (###)

(α1 + α2 + α3)^2 = 3(α1 α2 + α2 α3 + α3 α1).

Expressing (α1 α2 + α2 α3 + α3 α1) from (##) and putting here we get:

(α1 + α2 + α3)^2 = 3(6 + (α1 + α2 + α3))

Marking the sum (α1 + α2 + α3) with k we obtain:

k^2 - 3k - 18 = 0, and therefore:

k1 = -3 and k2 = 6

From (##) we find (α1 α2 + α1 α3 + α2 α3) and put into f(2) together with k:

a) k = k1 = -3

f(2) = 28

b) k = k2 = 6

f(2) = 10

Therefore, sum of all f(2) is 38. Ta da! :D

Step 1 : Restrict possibilities for $f(x)$ .

By conditions 3 and 4, $f(x) - (7x+2)$ vanishes at $x=0, 1$ so it is a multiple of $x(x-1)$ . Write $f(x) = h(x)x(x-1) + 7x+2$ for some linear polynomial $h(x)$ . Since $f(x)$ is monic, so is $h(x)$ . So $f(x) = (x-a)x(x-1) + 7x+2$ for some complex $a$ .

Step 2 : Use the second condition to deduce properties of $\alpha_i$ .

Write $g(x) = bx + c$ . Then $g(x_1) - g(x_2) = b(x_1 - x_2)$ . Hence:

$\alpha_3 - \alpha_2 = g(\alpha_2) - g(\alpha_1) = b(\alpha_2 - \alpha_1).$

By the same token, $\alpha_2 - \alpha_1 = b(\alpha_1 - \alpha_3)$ and $\alpha_1 - \alpha_3 = b(\alpha_3 - \alpha_2)$ so we have: $b^3(\alpha_3 - \alpha_2) = 1$ . If $b^3 \ne 0$ , then all the alpha's are equal, which is impossible by condition 1. Hence $b^3 = 1$ so $b=1$ or $b$ is a third root of unity.

In the first case, $b=1$ implies $\alpha_2 - \alpha_1 = \alpha_1 - \alpha_3$ and so $\alpha_1 + \alpha_2 + \alpha_3 = 3\alpha_1$ . By symmetry, this value also equals $3\alpha_2$ and $3\alpha_3$ and so all alphas are equal, which is impossible. Hence: $b = \omega$ is a third root of unity. Using the equality $\omega^2 + \omega + 1 = 0$ , we get:

$\alpha_2 - \alpha_1 = \omega(\alpha_1 - \alpha_3),\ \alpha_3 - \alpha_2 = \omega^2(\alpha_1 - \alpha_3).$

This gives:

$(\alpha_2 - \alpha_1)^2 + (\alpha_1 - \alpha_3)^2 + (\alpha_3 - \alpha_2)^2 = (1+\omega^2 + \omega^4)(\alpha_1 - \alpha_3)^2 = 0.$

Hence: $\alpha_1^2 + \alpha_2^2 + \alpha_3^2 = \alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1.$

Step 3 : Solve for $a$ .

We have: $f(x) = x^3 - (a+1)x^2 + (7+a)x + 2$ so the roots satisfy:

$\alpha_1 + \alpha_2 + \alpha_3 = a+1, \ \alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1 = a+7.$

From the conclusion at step 2, we get: $(a+1)^2 = 3(a+7)$ which gives $a^2 - a - 20 = 0$ . Hence, $a= 5, -4$ , which corresponds to $f(2) = 28, 10$ respectively.

[ Note : one also needs a certain amount of work to show that all these three steps are reversible, i.e. that $a=5, -4$ are both valid solutions. ]

Suppose $g(x)=mx+b$ . Note that $g(g(g(\alpha_i)))=\alpha_i$ for $i=1,2,3$ . In general, we have $g(g(g(x))) = m^3 x+m^2b+mb+b$ . Then we have $\alpha_i(1-m^3) = m^2b+mb+b$ , so if $m^3\neq 1$ we would have all roots identical, equal to $\frac{b}{1-m}$ . Then it must be that $m^3=1$ . $m=1$ forces $b=0$ , again making roots identical, so we must have $m$ a primitive third root of unity.

Suppose $r$ is a root of $f$ . Then the other roots are $mr+b$ and $m^2r+mb+b = m^2r-m^2b$ , and so $f(x) = (x-r)(x-m^2 r+m^2 b)(x-m r-b)$ . Note that if we had instead picked $g(x)=m^2 x-m^2 b$ , the inverse of our current choice, we would end up with the same set of roots for $f$ , and so the set of polynomials $f$ doesn't depend on which of the two third roots of unity $m$ is. Expanding, noting that $f(0)=2$ fixes the constant term as 2, we have $f(x) = x^3 -(m^2r+mr+r+b-m^2b)x^2 +(m^3r^2+m^2r^2+mr^2+rb-m^2 rb-m^3 rb+m^2 rb-m^2b^2)x+2$ . Or, simplifying, $f(x)= x^3+(m^2-1)bx^2-m^2b^2x+2$ , where we have used $m^3=1$ and $m^2+m+1=0$ .

Now, enforcing $f(1)=9$ gives $1+m^2b-b-m^2b^2+2=9$ , or $m^2b-b-m^2b^2=6$ . We have $f(2)=8+4(m^2b-b)-2m^2b^2+2$ . Substituting in the previous expression gives $f(2) = 10+2\cdot 6+2(m^2-1)b$ . We note that the $f(1)=0$ condition determines $b$ as the root of a quadratic, and the value of $f(2)$ depends only on the value of $b$ . Then the answer is $22+(m^2-1)b_1+22+(m^2-1)b_2$ for $b_i$ the possible values of b. The sum of values of $b$ is $\frac{m^2-1}{m^2} = 1-m$ by Vieta's formula. Then our answer is $44+2(m^2-1)(1-m) = 44+2(m^2+m-2) = 44-6 = 38$ .

First condition implies $f(x) = (x-\alpha_1)(x-\alpha_2)(x-\alpha_3).$ The condition 3 implies $f(0) = - \alpha_1 \alpha_2 \alpha_3 = 2. \qquad (\#1)$ The condition 4 implies, after using Eq.(#1), $\sum \alpha_1 \alpha_2 - \sum \alpha_i = 6, \qquad (\#2)$ where $\sum \alpha_1\alpha_2 \equiv \alpha_1\alpha_2 + \alpha_2\alpha_3 + \alpha_3\alpha_1$ .

The condition 2 implies that $g(x) = ax + b, \quad \text{and}$ $a \, \alpha_1 + b = \alpha_2; \quad a \, \alpha_2 + b = \alpha_3; \quad a \, \alpha_3 + b = \alpha_1.$ Solving for $a$ with the first two equations and the last two equation and matching them, after some algebra, gives $\sum \alpha_i^2 = \sum \alpha_1 \alpha_2 \quad \Rightarrow$ $\left( \sum \alpha_i \right)^2 = \sum \alpha_i^2 + 2\sum \alpha_1 \alpha_2 = 3\sum \alpha_1 \alpha_2.$ The same equation is obtained by doing the same for $b$ as well. Substituting this in equation (#2) gives, $\left( \sum \alpha_i \right)^2 - 3\sum \alpha_i - 18 = 0.$ Solving the above equation we find that the possibilities are $\sum \alpha_i = 6, \quad \sum \alpha_1 \alpha_2 = 12;$ and $\sum \alpha_i = -3, \quad \sum \alpha_1 \alpha_2 = 3.$ Now for $f(2)$ , after using Eq.(#1), $f(2) = 10 + 2 \sum \alpha_1 \alpha_2 - 4 \sum \alpha_i.$ Thus, the possible values of $f(2)$ are 10 and 28, whose sum is 38.

Let $f(x)= x^3 + ax^2 + bx + c$ . Since $f(0)= 2$ , $c= 2$ . Since $f(1)= 9$ , $a+b+c+1= 9 \implies a+b= 6$ . Hence, we have $f(x)= x^3 + ax^2 + (6-a)x + 2$ for some $a$ . We wish to determine all possible values for $a$ , as different values for $a$ produce different possibilities for $f(x)$ .

According to Vieta's formula, we have the following relations. $\alpha_1 + \alpha_2 + \alpha_3= -a$ $\alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1 = 6-a$ $\alpha_1 \alpha_2 \alpha_3 = -2$

Now let $g(x)= px+q$ , where $p,q$ are constants and $p \neq 0$ . According to the question, we have the following relations.
$p \alpha_1 + q= \alpha_2 \ .....(1)$ $p \alpha_2 + q= \alpha_3 \ .....(2)$ $p \alpha_3 + q= \alpha_1 \ .....(3)$

Solving for $p$ and $q$ in equations $(1)$ and $(2)$ yield
$p= \frac{\alpha_2 - \alpha_3}{\alpha_1-\alpha_2}$ $q= \frac{\alpha_1 \alpha_3 - \alpha_2^2}{\alpha_1-\alpha_2}$

Plugging these values in $(3)$ and multiplying both sides by $\alpha_1 - \alpha_2$ , we obtain $\alpha_1 \alpha_3 - \alpha_2^2 + \alpha_2 \alpha_3 - \alpha_3^2= \alpha_1^2 - \alpha_2 \alpha_2$ $\implies \alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1 = \alpha_1^2 + \alpha_2^2 + \alpha_3 ^2$

But we know that $\alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1 = 6-a$ $\alpha_1 ^2 + \alpha_2^2 + \alpha_3^2 = (\alpha_1 + \alpha_2 + \alpha_3)^2 - 2(\alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1) = a^2 - 2(6-a)$ .

Hence, we find out $6-a = a^2 - 2(6-a)$ . Solving yields $a= -6, +3$ , which gives the corresponding possibilities for $f(x)$ : $f(x)= x^3 - 6x^2 + 12x + 2$ $or$ $f(x)= x^3 + 3x^2 + 3x + 2$

Thus, the sum of all possible values of $f(2)$ is $2^3 - 6.2^2 + 12.2 + 2 + 2^3 + 3.2^2 + 3.2 + 2 = \boxed{38}$ .

Because $f$ is monic, $f(x) = (x-\alpha-1)(x-\alpha_2)(x-\alpha_3)$ Let $\displaystyle{A=\sum_{cyc}\alpha_1,\\B=\sum_{cyc}\alpha_1\alpha_2,\\C=\alpha_1\alpha_2\alpha_3}$ Substituting $f(0), f(1)$ gives us the following. $f(0)=-C=2\implies C=-2$ $f(1)=1-A+B-C=9\implies B=A+6$

To use the "cyclic root" condition, note that $g$ , a linear function, is an affine transformation; it rotates, dilates, and translates. Specifically, it has a constant angle of rotation. Since $g$ takes the line segment $\alpha_1\alpha_2$ to $\alpha_2\alpha_3$ , its angle of rotation is equal to the exterior angle at $\alpha_2$ . By symmetry, that angle of rotation is also equal to the exterior angles at $\alpha_3, \alpha_1$ . With 3 equal angles, the triangle formed by $\alpha_1,\alpha_2,\alpha_3$ is equilateral.

Lemma: For all equilateral triangles with vertices at complex numbers $x, y, z$ , we have the following identity: $x^2+y^2+z^2=xy+yz+zx$ Proof: $(z-x)=\pm\omega (y-x)$ where $\omega$ is a primitive sixth root of unity. Specifically, $\omega^2-\omega+1=0$ Then $\left(\frac{z-x}{y-x}\right)^2-\left(\frac{z-x}{y-x}\right)+1=0\\ (z-x)^2-(z-x)(y-x)+(y-x)^2=0\\ x^2+y^2+z^2=xy+yz+zx$

Using this identity, we can plug in this "cyclic roots" condition easily. $\displaystyle{\sum_{cyc}\alpha_1^2=B\\ A^2-2B=B\\ A^2=3B}$

We substitute $B=A+6$ to obtain a quadratic in $A$ . $A^2=3(A+6)\\ A^2-3A-18=0$ There are two roots for $A$ , and their sum is $3$ .

We wish to find the sum of possible values of $f(2)$ . $f(2)=8-4A+2B-C=8-4A+2(A+6)-(-2)=22-2A$ The sum of the possible values of $f(2)$ (there are 2 of them) is then simply $22\cdot2 - 2(3) = \boxed{38}$ .