Let $\alpha=\frac{-1+i \sqrt{3}}{2}$ and $\beta=\frac{-1-i \sqrt{3}}{2}, i=\sqrt{-1}$. If $(7-7 \alpha+9 \beta)^{20}+(9+7 \alpha-7 \beta)^{20}+(-7+9 \alpha+7 \beta)^{20}+(14+7 \alpha+7 \beta)^{20}=\mathrm{m}^{10}$, then m is

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Complex Numbers

Let

α = \frac{- 1 + i 3}{2}

and

β = \frac{- 1 - i 3}{2}, i = - 1

. If

(7 - 7 α + 9 β)^{20} + (9 + 7 α - 7 β)^{20} + (- 7 + 9 α + 7 β)^{20} + (14 + 7 α + 7 β)^{20} = m^{10}

, then m is

Answer: 49

With

α = ω

and

β = ω^{2}

the primitive cube roots of unity, the four bases are

z_{1} = 7 - 7 ω + 9 ω^{2}

z_{2} = 9 + 7 ω - 7 ω^{2}

z_{3} = - 7 + 9 ω + 7 ω^{2}

and

z_{4} = 14 + 7 (ω + ω^{2}) = 14 - 7 = 7

. Using

ω^{2} = - 1 - ω

gives

z_{1} = 6 - 83 i

, and a direct check shows

z_{2} = ω z_{1}

and

z_{3} = ω^{2} z_{1}

. Since

ω^{20} = ω^{2}

and

ω^{40} = ω

z_{1}^{20} + z_{2}^{20} + z_{3}^{20} = z_{1}^{20} (1 + ω^{20} + ω^{40}) = z_{1}^{20} (1 + ω^{2} + ω) = 0.

The sum therefore reduces to

z_{4}^{20} = 7^{20} = (49)^{10}

, so

m = 49

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Integer answer