Given $x^2 + x + 1 = 0$, the roots are $x = \omega$ and $x = \omega^2$, where $\omega$ is a complex cube root of unity.

We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Also, $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\frac{1}{\omega} = \omega^2$, and $\frac{1}{\omega^2} = \omega$.

Consider the term $x^n + \frac{1}{x^n}$. If $x = \omega$, then $x^n + \frac{1}{x^n} = \omega^n + \frac{1}{\omega^n} = \omega^n + \omega^{-n} = \omega^n + \omega^{2n}$.

Now, we analyze the values of $\omega^n + \omega^{2n}$ for different values of $n$:

• If $n$ is a multiple of 3, i.e., $n = 3k$, then $\omega^{3k} + \omega^{6k} = 1 + 1 = 2$.
• If $n = 3k + 1$, then $\omega^{3k+1} + \omega^{6k+2} = \omega + \omega^2 = -1$.
• If $n = 3k + 2$, then $\omega^{3k+2} + \omega^{6k+4} = \omega^2 + \omega^4 = \omega^2 + \omega = -1$.

Therefore, $x^n + \frac{1}{x^n} = \begin{cases} 2, & \text{if } n \equiv 0 \pmod{3} \\ -1, & \text{if } n \not\equiv 0 \pmod{3} \end{cases}$

We need to evaluate $\sum_{n=1}^{25} (x^n + \frac{1}{x^n})^4$.

Since $(x^n + \frac{1}{x^n})^4 = \begin{cases} 2^4 = 16, & \text{if } n \equiv 0 \pmod{3} \\ (-1)^4 = 1, & \text{if } n \not\equiv 0 \pmod{3} \end{cases}$

The values of $n$ from 1 to 25 that are multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24. There are 8 such values.

The remaining $25 - 8 = 17$ values are not multiples of 3.

Therefore, the sum is $8 \cdot 16 + 17 \cdot 1 = 128 + 17 = 145$.

Given the equation $x^2 + x + 1 = 0$, the roots are $\omega$ and $\omega^2$, where $\omega$ is a complex cube root of unity.

We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Also, $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$.

Consider the term $x^n + \frac{1}{x^n}$. If $x = \omega$, then $x^n + \frac{1}{x^n} = \omega^n + \frac{1}{\omega^n} = \omega^n + \omega^{-n} = \omega^n + \omega^{2n}$.

Now, we analyze the values of $\omega^n + \omega^{2n}$ for different values of $n$:

• If $n$ is a multiple of 3, i.e., $n = 3k$, then $\omega^{3k} + \omega^{6k} = 1 + 1 = 2$.
• If $n = 3k + 1$, then $\omega^{3k+1} + \omega^{6k+2} = \omega + \omega^2 = -1$.
• If $n = 3k + 2$, then $\omega^{3k+2} + \omega^{6k+4} = \omega^2 + \omega^4 = \omega^2 + \omega = -1$.

Therefore, $x^n + \frac{1}{x^n} = \begin{cases} 2, & \text{if } n \equiv 0 \pmod{3} \\ -1, & \text{if } n \not\equiv 0 \pmod{3} \end{cases}$

We need to evaluate $\sum_{n=1}^{25} (x^n + \frac{1}{x^n})^4$.

Since $(x^n + \frac{1}{x^n})^4 = \begin{cases} 2^4 = 16, & \text{if } n \equiv 0 \pmod{3} \\ (-1)^4 = 1, & \text{if } n \not\equiv 0 \pmod{3} \end{cases}$

The values of $n$ from 1 to 25 that are multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24. There are 8 such values.

The remaining $25 - 8 = 17$ values are not multiples of 3.

Therefore, the sum is $8 \cdot 16 + 17 \cdot 1 = 128 + 17 = 145$.