First, recognize that $\alpha$ and $\beta$ are cube roots of unity. Specifically, $\alpha$ and $\beta$ satisfy the equation $x^2 + x + 1 = 0$, since one of their sum is $-1$ and their product is 1. In terms of cube roots of unity, $\alpha$ is $\omega$ and $\beta$ is $\omega^2$, where $\omega = e^{2\pi i/3}$. These satisfy $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Next, compute:
- For $A = 7 - 7\alpha + 9\beta$:
- $7 - 7\alpha + 9\beta = 7 - 7\omega + 9\omega^2$
- Using $1 + \omega + \omega^2 = 0$, rewrite $7 - 7(\omega - 1) + 9(\omega^2 - 1) = 7 + 2 = 21$
- For $B = 9 + 7\alpha - 7\beta$:
- $9 + 7\omega - 7\omega^2 = 9 + 7(\omega - \omega^2) \to 9 + 7(-1) = 2$
- For $C = -7 + 9\alpha + 7\beta$:
- $-7 + 9\omega + 7\omega^2 = -7 + 9(-\omega^2 - 1) + 7$ from earlier
- Thus, rewrite as $-7 + 16 = 9
- For $D = 14 + 7\alpha + 7\beta$:
- $14 + 7\omega + 7\omega^2 = 14 + 7(-1) = 7
Compute the sum of each computed value to the power of 20:
$A^{20} + B^{20} + C^{20} + D^{20} = 21^{20} + 2^{20} + 9^{20} + 7^{20}$. Recognize complex terms are unit root rotations changing cyclically every third power multiplied with unity j-th roots characters, rendering computations inefficiently large, simplifying down to integers solely influenced by unity roots, concluding each path internally rooted as 1 or sum $(1)^{20}$.
This expression equated to $m^{10}$ simplifies naturally to overall unity schema, reaching an LCM effector of 49 unit harmonic multipliers.
Similar Questions
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Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
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- Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $\det(Q) = 2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:
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Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $\det(Q) = 2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:
