If x^ 2 +x+1=0 , then the value of (x+(1)/(x))^ 4 +(x^ 2 + 1 x^ 2 )^ 4 +(x^ 3 + 1 x^ 3 )^ 4 + +(x^ 25 + 1 x^ 25 )^ 4 is:

Correct answer: C. 145 Given x^2 + x + 1 = 0 , so x = ω (primitive cube root of unity) with x^3 = 1 . From the equation: x + (1)/(x) = -1 . Since x^3 = 1 , value of x^n + x^ -n has period 3: n 0 3 : x^n + x^ -n = 2 n 1, 2 3 : x^n + x^ -n = -1 Fourth powers: (2)^4 = 16 , (-1)^4 = 1 . For n = 1 to 25 : 8 multiples of 3 contribute 8 × 16 = 128 , remaining 17 terms contribute 17 × 1 = 17 . Total = 128 + 17 = 145 .

JEE Main 202621 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Complex Number

If x^ 2 +x+1=0 , then the value of (x+(1)/(x))^ 4 +(x^ 2 + 1 x^ 2...

x^{2}+x+1=0

, then the value of \left(x+\frac{1}{x}\right)^{4}+\left(x^{2}+\frac{1}{x^{2}}\right)^{4}

+\left(x^{3}+\frac{1}{x^{3}}\right)^{4}

+\ldots+\left(x^{25}+\frac{1}{x^{25}}\right)^{4} is:

Options

A162
B175
C145
D128

Correct answer

C. 145

Step-by-step solution

Given

x^2 + x + 1 = 0

, so

x = \omega

(primitive cube root of unity) with

x^3 = 1

.
From the equation:

x + \frac{1}{x} = -1

.
Since

x^3 = 1

, value of

x^n + x^{-n}

has period 3:

n \equiv 0 \pmod{3}

x^n + x^{-n} = 2

n \equiv 1, 2 \pmod{3}

x^n + x^{-n} = -1

Fourth powers:

(2)^4 = 16

(-1)^4 = 1

.
For

n = 1

25

: 8 multiples of 3 contribute

8 \times 16 = 128

, remaining 17 terms contribute

17 \times 1 = 17

.
Total

= 128 + 17 = 145

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