2018 welcomes freaky Polynomials!

Since $\alpha$ , $\beta$ and $\gamma$ are roots of $x^3-x^2-1=0$ , by Vieta's formula , we have:

$\begin{cases} \alpha + \beta + \gamma = 1 \\ \alpha \beta + \beta \gamma + \gamma \alpha = 0 \\ \alpha \beta \gamma = 1 \end{cases}$

And that:

$\begin{cases} \alpha^3 - \alpha^2 - 1 & = 0 & \implies \alpha^3 = \alpha^2 + 1, & \alpha^4 = \alpha^2 + \alpha + 1 \\ \beta^3 - \beta^2 - 1 & = 0 & \implies \beta^3 = \beta^2 + 1, & \beta^4 = \beta^2 + \beta + 1 \\ \gamma^3 - \gamma^2 - 1 & = 0 & \implies \gamma^3 = \gamma^2 + 1, & \gamma^4 = \gamma^2 + \gamma + 1 \end{cases}$

Then the roots of $x^3-ax^2 + bx + c =0$ are:

$\begin{cases} \alpha^2 + \beta^3 + \gamma^4 = \alpha^2 + \beta^2 + 1 + \gamma^2 + \gamma + 1 = \gamma + 3 \\ \beta^2 + \gamma^3 + \alpha^4 = \alpha + 3 \\ \gamma^2 + \alpha^3 + \beta^4 = \beta + 3 \end{cases}$

By Vieta's formula again:

$\begin{aligned} a & = \alpha + 3 + \beta +3 + \gamma + 3 = 10 \\ b & = (\alpha + 3)(\beta +3)+ (\beta +3)(\gamma + 3) +(\gamma + 3)(\alpha + 3) \\ & = \alpha \beta + \beta \gamma + \gamma \alpha + 6(\alpha + \beta + \gamma) + 27 \\ & = 0+6(1) + 27 = 33 \\ c & = -(\alpha + 3)(\beta +3)(\gamma + 3) \\ & = - \alpha \beta \gamma - 3(\alpha \beta + \beta \gamma + \gamma \alpha) - 9(\alpha + \beta + \gamma) - 27 \\ & = - 1 - 3(0) - 9(1) - 27 = - 37 \end{aligned}$

$\implies a+b+c = 10+33-37 = \boxed{6}$

By Vieta's formulas, we have: $\alpha+\beta+\gamma=1$ , $\alpha\beta+\alpha\gamma+\beta\gamma=0$ and $\alpha\beta\gamma=1$ . Let $x_0$ be a root of $x^3-x^2-1=0$ , then $x_0^3=x_0^2+1$ and $x_0^4=x_0^3+x_0=x_0^2+x_0+1$ .

Let $r_1=\alpha^2 + { \beta }^{ 3 } + { \gamma }^{ 4 }$ , $r_2={ \beta }^{ 2 } + { \gamma }^{ 3 } + { \alpha }^{ 4 }$ and $r_3={ \gamma }^{ 2 } + \alpha ^{ 3 } + \beta ^{ 4 }$ the new roots, then $r_1=\alpha^2+\beta^2+\gamma^2+\gamma+2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma)+\gamma+2=\gamma+3$ , similarly $r_2=\alpha+3$ and $r_3=\beta+3$ .

We see that the new roots are the old roots increased by 3, so let's do the change $y=x+3 \implies x=y-3$ and substitute: $(y-3)^3-(y-3)^2-1=0 \implies y^3-10y^2+33y-37=0$ .

Finally, we get $a=10$ , $b=33$ and $c=-37$ , making the answer $10+33-37=\boxed{6}$ .