If x^2 - x + 1 = 0 , then what is (x - (1)/(x))^2 + (x - (1)/(x))^4...

Given $x^2 - x + 1 = 0$

Dividing the entire equation by $x$, we get:

$x - 1 + \dfrac{1}{x} = 0 \Rightarrow x + \dfrac{1}{x} = 1$

Using the algebraic identity \left(x - \dfrac{1}{x}\right)^2 = \left(x + \dfrac{1}{x}\right)^2 - 4$<br><br>Substituting the value of$x + \dfrac{1}{x}$:<br><br>\left(x - \dfrac{1}{x}\right)^2 = (1)^2 - 4 = -3$

The given expression is \left(x - \dfrac{1}{x}\right)^2 + \left(x - \dfrac{1}{x}\right)^4 + \left(x - \dfrac{1}{x}\right)^8$<br><br>This can be written as \left(x - \dfrac{1}{x}\right)^2 + \left(\left(x - \dfrac{1}{x}\right)^2\right)^2 + \left(\left(x - \dfrac{1}{x}\right)^2\right)^4$

Substituting \left(x - \dfrac{1}{x}\right)^2 = -3$into the expression:<br><br>$(-3) + (-3)^2 + (-3)^4$<br><br>$= -3 + 9 + 81$<br><br>$= 87$<br><br>Answer:$87