If α satisfies the equation x²+x+1=0 and (1+α)⁷=A+B α+C α², A, B, C ≥ 0, then 5(3 A-2 B-C) is equal to ____________.

Solution

Maths sir · 4 steps

Swap 1+α for -α² first; then the seventh power becomes one clean term.

1. Roots are ω type. The roots of x²+x+1=0 are the non-real cube roots of unity, so α³ = 1 and 1+α+α² = 0. That gives 1+α = -α².

   (1+α)⁷ = (-α²)⁷ = -α¹⁴ = -α¹²·α² = -α²

2. Make coefficients non-negative. -α² has a negative coefficient, but the question needs A, B, C ≥ 0. Use -α² = 1+α, so A = 1, B = 1, C = 0.
3. Check other forms. You may add any k(1+α+α²) = 0 with k ≥ 0, giving A = B = 1+k, C = k. Then 3A-2B-C = 3+3k-2-2k-k = 1, the same value every time.
4. So. 5(3A-2B-C) = 5 × 1 = 5.

The common slip: -5. Dropping the minus sign in (-α²)⁷ gives α², so C = 1 and the answer comes out as -5.

The idea to keep. With cube roots of unity, rewrite 1+ω as -ω² before raising powers, then cut exponents using ω³ = 1. Use 1+ω+ω² = 0 to move signs.

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