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UPSC NDA I 2025 Mathematics Paper

Question : 14 of 120

Marks: +1, -0

x^2-x+1=0

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

Solution:

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} \; \; \; \text{ or } \; \; \; 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

\therefore

The value of the expression is 87 .

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