If x² + x + 1 = 0, then

(x + 1/x)⁴ + (x² + 1/x²)⁴ + (x³ + 1/x³)⁴ + ... + (x²⁵ + 1/x²⁵)⁴   is

The given equation is:
$x^2+x+1=0$
This equation represents the quadratic equation whose roots are the non-real cube roots of unity, commonly denoted by $\omega$ and ${\omega }^2$.

Recall the fundamental properties of the cube root of unity:
1. ${\omega }^3=1$
2. $1+\omega +{\omega }^2=0$ (which implies $\omega +{\omega }^2=-1$)
3. $\frac{1}{\omega }={\omega }^2$ and $\frac{1}{{\omega }^2}=\omega$

Let $x=\omega$. We want to find the value of the expression:
$S=\sum _{k=1}^{25}{(x^k+\frac{1}{x^k})}^4$

Let's analyze the term $x^k+\frac{1}{x^k}$ based on the value of $k$ modulo 3:

Case 1: When $k$ is a multiple of 3 (i.e., $k=3m$ for an integer $m$)
Since $x^3={\omega }^3=1$, we have:
$x^k+\frac{1}{x^k}={\left(x^3\right)}^m+\frac{1}{{\left(x^3\right)}^m}=1^m+\frac{1}{1^m}=1+1=2$
Therefore:
${(x^k+\frac{1}{x^k})}^4=2^4=16$

Case 2: When $k$ is not a multiple of 3 (i.e., $k=3m+1$ or $k=3m+2$)
If $k=3m+1$:
$x^k+\frac{1}{x^k}=x+\frac{1}{x}=\omega +{\omega }^2=-1$
If $k=3m+2$:
$x^k+\frac{1}{x^k}=x^2+\frac{1}{x^2}={\omega }^2+\omega =-1$
In both cases, we get:
${(x^k+\frac{1}{x^k})}^4={(-1)}^4=1$

Now, we count the number of terms of each type for $k$ from 1 to 25:
- The multiples of 3 in the range $[1,25]$ are: $3,6,9,12,15,18,21,24$.
The number of such terms is $8$ (since $25=3\times 8+1$).
- The remaining terms are not multiples of 3.
The number of such terms is $25-8=17$.

Calculating the total sum $S$:
$S=(8\times 16)+(17\times 1)$
$S=128+17=145$

Thus, the value of the given expression is 145.