Question

If \( x^2 + x + 1 = 0 \), then evaluate \[ \left(x + \frac{1}{x}\right)^4 + \left(x^2 + \frac{1}{x^2}\right)^4 + \left(x^3 + \frac{1}{x^3}\right)^4 + \cdots + \left(x^{25} + \frac{1}{x^{25}}\right)^4. \]

Hint:

For equations involving roots of unity:
Use periodicity: \( x^3 = 1 \)
Values of \( x^n + \frac{1}{x^n} \) repeat every 3
Count terms carefully in long summations

Answer 1

Correct Answer: 145

Step 1: From the given equation: \[ x^2 + x + 1 = 0 \Rightarrow x^3 = 1,\; x \neq 1 \] Thus, \( x \) is a complex cube root of unity.
Step 2: Compute the basic value: \[ x + \frac{1}{x} = -1 \]
Step 3: Use periodicity: \[ x^n + \frac{1}{x^n} = \begin{cases} 2, & \text{if } 3 \mid n \\ -1, & \text{if } 3 \nmid n \end{cases} \]
Step 4: Raise to the 4th power: \[ (-1)^4 = 1, 2^4 = 16 \]
Step 5: Count terms from \( n = 1 \) to \( 25 \): Multiples of 3: \[ \left\lfloor \frac{25}{3} \right\rfloor = 8 \] Non-multiples of 3: \[ 25 - 8 = 17 \]
Step 6: Compute the sum: \[ 8 \times 16 + 17 \times 1 = 128 + 17 = 145 \]