Q163 / 269 · NDA 2025 (Phase 1) · Mathematics

If $x^2 - x + 1 = 0$, then what is $\left(x - \dfrac{1}{x}\right)^2 + \left(x - \dfrac{1}{x}\right)^4 + \left(x - \dfrac{1}{x}\right)^8$ equal to?

  • A
    $81$
  • B
    $85$
  • Correct answer
    C
    $87$
  • D
    $90$

Solution

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 $\left(x - \dfrac{1}{x}\right)^2 = \left(x + \dfrac{1}{x}\right)^2 - 4$

Substituting the value of $x + \dfrac{1}{x}$:

$\left(x - \dfrac{1}{x}\right)^2 = (1)^2 - 4 = -3$

The given expression is $\left(x - \dfrac{1}{x}\right)^2 + \left(x - \dfrac{1}{x}\right)^4 + \left(x - \dfrac{1}{x}\right)^8$

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:

$(-3) + (-3)^2 + (-3)^4$

$= -3 + 9 + 81$

$= 87$

Answer: $87$

Hence, the correct answer is option: 'C'.

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