By signing in you confirm you are 18 or older, or have your parent’s or guardian’s consent.

NDA 2 2024 Maths, question 4

Topic: Algebra

Q4

If ω≠1\omega\ne 1 is a cube root of unity, then what is (1+ω−ω2)100+(1−ω+ω2)100(1+\omega-\omega^{2})^{100}+(1-\omega+\omega^{2})^{100} equal to?

  1. (a) 2100ω22^{100}\omega^{2}
  2. (b) 2100ω2^{100}\omega
  3. (c) 21002^{100}
  4. (d) −2100-2^{100}Correct answer

Solution.

Answer: (d) −2100-2^{100}

  1. Using 1+ω+ω²=0, the two bases become −2ω² and −2ω.
  2. Their 100th powers sum to 2¹⁰⁰(ω²⁰⁰+ω¹⁰⁰)=2¹⁰⁰(ω²+ω)=−2¹⁰⁰.
  3. The exponents were reduced modulo 3.