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NDA 1 2025 Maths, question 14

Topic: Algebra

Q14

If x2−x+1=0x^{2} - x + 1 = 0, then what is (x−1x)2+(x−1x)4+(x−1x)8\left( x - \frac{1}{x} \right)^{2} + \left( x - \frac{1}{x} \right)^{4} + \left( x - \frac{1}{x} \right)^{8} equal to?

  1. (a) 81
  2. (b) 85
  3. (c) 87Correct answer
  4. (d) 90

Solution.

Answer: (c) 87

  1. Since x≠0x \neq 0, division by xx gives x+1/x=1x + 1/x = 1.
  2. Hence (x−1/x)2=(x+1/x)2−4=−3(x - 1/x)^{2} = (x + 1/x)^{2} - 4 = - 3.
  3. The required sum is −3+(−3)2+(−3)4=−3+9+81=87- 3 + ( - 3)^{2} + ( - 3)^{4} = - 3 + 9 + 81 = 87.