If $x{}^2-x+1=0$, then what is \((x-\frac{1}{x})^2+(x-\frac{1}{x})^4+(x-\frac{1}{x})^8\)

Calculation:

Given,

The equation is \( x^2 - x + 1 = 0 \)

We need to find the value of the following expression:

\( \left( x - \frac{1}{x} \right)^2 + \left( x - \frac{1}{x} \right)^4 + \left( x - \frac{1}{x} \right)^8 \)

The equation \( x^2 - x + 1 = 0 \) is solved as follows:

\( x = \frac{1 \pm \sqrt{-3}}{2} = e^{i \pi / 3} \quad \text{or} \quad x = e^{-i \pi / 3} \)

Now, substitute the value of \( x \) into the expression \( x - \frac{1}{x} \):

\( x - \frac{1}{x} = i\sqrt{3} \)

Evaluate the powers of \( x - \frac{1}{x} \)

Now, let's evaluate each term in the expression:

\( \left( x - \frac{1}{x} \right)^2 = (i \sqrt{3})^2 = -3 \)

\( \left( x - \frac{1}{x} \right)^4 = (-3)^2 = 9 \)

\( \left( x - \frac{1}{x} \right)^8 = 9^2 = 81 \)

Now, sum the values:

\( -3 + 9 + 81 = 87 \)

∴ The value of the expression is 87.

Hence, the correct answer is Option 3.