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JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Complex Numbers

If

x^{2} + x + 1 = 0

, then the value of

(x + \frac{1}{x})^{4} + (x^{2} + \frac{1}{x ^{2}})^{4} + (x^{3} + \frac{1}{x ^{3}})^{4} + \dots + (x^{25} + \frac{1}{x ^{25}})^{4}

is :

(A)128
(B)162
(C)175
(D)145✓ Correct

Answer: (D)

Solution

Since

x^{2} + x + 1 = 0

,

x

is a non-real cube root of unity, so

x^{3} = 1

and

x^{n} + x^{- n}

equals

2

when

3 ∣ n

and

- 1

otherwise. Thus

(x^{n} + x^{- n})^{4} = 16

for the eight multiples of

3

in

1, \dots, 25

and

1

for the remaining

17

values. The sum is

8 \cdot 16 + 17 \cdot 1 = 145

.

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option

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