Raised to the Power of 27!

Let x=alpha. y=beta z=gamma

Then ATQ.

x+y+z=0

xy+yz+zx=3

xyz=-9

From above 3 we can see that x^3+y^3+z^3=3xyz =-27 —————(1)

(xy)^3+(yz)^3+(zx)^3. =270 ——————(2)

x^9+y^9+z^9=-27(729-810)-2187=0 ————(3)

From the above result we can see that x^27+y^27+z^27=3(xyz)^9 =-1162261467

The problem can be solved using Newton Sums method.

Let $S_1 = \alpha + \beta + \gamma$ , $S_2=S_1 = \alpha\beta + \beta \gamma + \gamma \alpha$ , $S_3 = \alpha \beta \gamma$ and $P_n = \alpha^n + \beta^n + \gamma^n$ , where $n = 1,2,3...$ . Then

$\begin{cases} P_1 = S_1 \\ P_2 = S_1P_1 - 2S_2 \\ P_3 = S_1P_2 - S_2P_1 + 3S_3 \\ P_4 = S_1P_3 - S_2P_2 + S_3P_1 \\ ... \\ P_n = S_1P_{n-1} - S_2P_{n-2} + S_3P_{n-3} \end{cases}$

Using a spreadsheet, $P_{27} = \alpha^{27} + \beta^{27} + \gamma^{27}$ is readily calculated and it is $\boxed{-1162261467}$

Let $a,b,c$ be the roots of the equation, and

$\begin{aligned}S_n &= a^n + b^n + c^n \\ P_n &= (ab)^n + (bc)^n + (ca)^n \\ Q_n &= (abc)^n\end{aligned}$

From the equation, we have

$\begin{aligned} S_1 &= 0 \\ P_1 &= 3 \\ Q_1 &= -9 \end{aligned}$

Instead of using Newton's, I'll be using a beautiful algebraic expression

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

In terms of $S_n, P_n, Q_n$ , it will look like this

$\begin{aligned}S_{3n}-3Q_n&=S_n(S_{2n} - P_n) \\ P_{3n} - 3Q_{2n} &= P_n(P_{2n}-Q_nS_n) \end{aligned}$

The other two useful formula is

$\begin{aligned}S_n^2 &= S_{2n} + 2P_n \\ P_n^2 &= P_{2n} + 2S_nP_n\end{aligned}$

For $S_1 = 0$ , we can quickly obtain

$S_3 = 3Q_1 = -27$

By the same method,

$\begin{aligned} S_9 - 3Q_3 &= S_3(S_6-P_3) \\ \end{aligned}$

We need to solve for $S_6, P_3$ ,

$S_6 = S_3^2-2P_3$

$P_3 = P_1P_2 + 3Q_2$

Then, solve for $P_2$ ,

$P_2 = P_1^2-2S_1Q_1 = P_1^2$

Substitute to the equation, we have

$\begin{aligned} S_9 &= S_3[S_3^2-3(P_1^3+3Q_2)] +3Q_3\\&=0 \end{aligned}$

For $S_9 = 0$ , then we have

$\begin{aligned} S_{27}-3Q_9&=0 \\ S_{27} &= 3Q_9 \\&= -1162261467\end{aligned}$

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I think there's a faster solution, since I've used the calculator two times.