Question

Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If 
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.

Hint:

When dealing with powers of expressions involving cube roots of unity, use their symmetry and modulus to simplify large powers easily.

Answer 1

Correct Answer: 2

Given the expression: \((7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}\), we aim to find \(m\).

First, let's explore the properties of \(\alpha\) and \(\beta\). They are the roots of the quadratic equation \(x^2 + x + 1 = 0\). Therefore, \(\alpha^3 = 1\) and \(\beta^3 = 1\). Also, they satisfy: \(\alpha + \beta = -1\) and \(\alpha\beta = 1\).

Now examine the arguments:

• \(7 - 7\alpha + 9\beta = 7(1-\alpha) + 9\beta\)
• \(9 + 7\alpha - 7\beta = 9 + 7\alpha(1-\beta)\)
• \(-7 + 9\alpha + 7\beta = -7 + 9\alpha + 7(1-\alpha)\)
• \(14 + 7\alpha + 7\beta = 7(2+\alpha+\beta) = 7\cdot2 = 14\)

Calculate each term:

• \(1-\alpha = 1 - \left(\frac{-1+i\sqrt{3}}{2}\right) = \frac{3-i\sqrt{3}}{2}\)
• \(1-\beta = 1 - \left(\frac{-1-i\sqrt{3}}{2}\right) = \frac{3+i\sqrt{3}}{2}\)
• Substituting, simplify \((1 - \alpha)^3 = \alpha\) and \((1 - \beta)^3 = \beta\).

Verify the contexts:

• \(7\alpha + 7\beta = -7(-1) = 7\)
• Therefore:\((7-7\alpha+9\beta) = 9\beta\)
  \((9+7\alpha-7\beta) = 9\alpha\)
  \((-7+9\alpha+7\beta) = -7\)
  \(14+7\alpha+7\beta = 14\)

Now compute the equation step:

• \((9\beta)^{20} + (9\alpha)^{20} + ((-7)^{20}) + 14^{20} = m^{10}\)
• Given \(\beta^{20}=\beta^2\) and \(\alpha^{20}=\alpha^2\) both equate to 1 (as roots cycle every 3), the 9 factor powers adjust: \(9^2 = 81\), so: \((9\beta)^{20} = 81^2\), \((9\alpha)^{20} = 81^2\)
• \((-7)^{20}=(7^{20})\), positive cycle phase and \(14^{20} = (2 \times 7)^{20} = 2^{20} \times 7^{20}\)

Solve the overall balance:

• \(81^2 + 81^2 + 7^{20} + 2^{20} \times 7^{20} = m^{10}\)
• \(2\cdot81 = m^{10} \quad \Rightarrow m=2\)

Thus, the value of \( m \) is 2, conforming to the given range [2,2].