Let z = (1 - i√3)/2, i = √(- 1). Then the value of 21 + ( z + 1/z )³ + ( z² + 1/z² )³ + ( z³ + 1/z³ )³ + .... + ( z²¹ + 1/z²¹ )³ is ______________.

Solution

Maths sir · 3 steps

Spot that z = -ω. Then every bracket takes only a few values, and they repeat every 6 terms.

1. Recognise z. With ω = (-1 + i√3)/2, we get z = (1 - i√3)/2 = -ω. Use ω³ = 1, 1/ω = ω² and 1 + ω + ω² = 0.

   zᵏ + 1/zᵏ = (-1)ᵏ(ωᵏ + ω²ᵏ) = (-1)ᵏ × 2, 3 | k ; (-1)ᵏ × (-1), 3 k

   k (mod 6)	zᵏ + 1/zᵏ	Cube
   1	1	1
   2	-1	-1
   3	-2	-8
   4	-1	-1
   5	1	1
   6	2	8

2. Group by sixes. One full cycle of 6 cubes adds to 1 - 1 - 8 - 1 + 1 + 8 = 0. Since 21 = 18 + 3, the first 18 cubes add to 0. The last three, k = 19, 20, 21, behave like k = 1, 2, 3 and give 1 - 1 - 8 = -8.
3. So. The value is 21 + (-8) = 13.

The common slip: 63. Treating z as ω and dropping the minus sign makes the cubes -1, -1, 8 repeat every 3 terms. That gives 21 + 7 × 6 = 63, not 13.

The idea to keep. For a power sum with (± 1 ± i√3)/2, first write it as ±ω or ±ω². Then find the period of the terms, cancel the full cycles and work out only the leftover terms.

Answer from the NTA final answer key. Solution by PrepWiser.