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NDA 1 2024 Maths, question 14

Topic: Algebra

Q14

If ω≠1\omega\ne 1 is a cube root of unity, then what are the solutions of (z−100)3+1000=0(z-100)^{3}+1000=0 ?

  1. (a) 10(1−ω),10(10−ω2),10010(1-\omega),10(10-\omega^{2}),100
  2. (b) 10(10−ω),10(10−ω2),9010(10-\omega),10(10-\omega^{2}),90Correct answer
  3. (c) 10(1−ω),10(10−ω2),100010(1-\omega),10(10-\omega^{2}),1000
  4. (d) (1+ω),(10+ω2),−1(1+\omega),(10+\omega^{2}),-1

Solution.

Answer: (b) 10(10−ω),10(10−ω2),9010(10-\omega),10(10-\omega^{2}),90

  1. We need (z−100)³=−1000.
  2. The three cube roots on the right are −10, −10ω and −10ω².
  3. Adding 100 gives 90, 10(10−ω) and 10(10−ω²), which are the roots in (b).