If $𝑥^2+𝑥+1=0$, then the value of $\left(𝑥+\frac{1}{𝑥}\right)^4+\left(𝑥^2+\frac{1}{𝑥^2}\right)^4+\left(𝑥^3+\frac{1}{𝑥^3}\right)^4+\ldots+\left(𝑥^{25}+\frac{1}{𝑥^{25}}\right)^4$ is:

Question

Accepted Answer

B. 1. Recognize that $𝑥^2+𝑥+1=0$ means $𝑥$ is a non-real **cube root of unity** $\omega$, since $\omega$ satisfies exactly this equation.

2. Recall the key property of cube roots of unity: $\omega^k+\frac{1}{\omega^k}$ depends on whether $𝑘$ is a multiple of 3.
$$k=3n \ \Rightarrow\ \omega^k+\frac{1}{\omega^k}=2, \qquad k
e3n \ \Rightarrow\ \omega^k+\frac{1}{\omega^k}=-1$$

3. Among $k=1$ to $25$, count how many are multiples of 3 and how many are not.
$$	ext{Multiples of 3 } (k=3,6,\dots,24): 8 	ext{ values}$$
$$	ext{Remaining: } 25-8=17 	ext{ values}$$

4. Raise each case to the 4th power, since the sum required is of 4th powers: multiples of 3 contribute $2^4=16$ each, others contribute $(-1)^4=1$ each.
$$	ext{Sum} = 8	imes16 + 17	imes1$$

5. Compute the total.
$$= 128+17 = 145$$

Hence, the answer is **145**, option **B**.