Question
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:
JEE Main
Mathematics
Complex Number
2026
JEE Main 2026 (21 January Shift 1)
JEE Main Mathematics Question (2026) — Solution
Options
- A. 162
- B. 175
- C. 145
- D. 128
Answer
C. 145
Step-by-step solution
Given x^2 + x + 1 = 0, so x = (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.
Related: Mathematics — Complex Number · All PYQ Banks