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

$$x^2 + x + 1 = 0$$ means $$x = \omega$$ or $$\omega^2$$ (cube roots of unity).

For $$x = \omega$$: $$x^n + \frac{1}{x^n} = \omega^n + \omega^{-n} = \omega^n + \omega^{2n}$$ (since $$\omega^{-1} = \omega^2$$).

If $$n \equiv 0 \pmod{3}$$: $$\omega^n + \omega^{2n} = 1 + 1 = 2$$, so $$(x^n + 1/x^n)^4 = 16$$.

If $$n \not\equiv 0 \pmod{3}$$: $$\omega^n + \omega^{2n} = -1$$, so $$(x^n + 1/x^n)^4 = 1$$.

From n=1 to 25: multiples of 3 are 3,6,9,12,15,18,21,24 → 8 values.

Non-multiples: 25 - 8 = 17 values.

Sum = $$8 \times 16 + 17 \times 1 = 128 + 17 = 145$$.

The answer is Option 4: 145.