Complex Numbers
Cube Root of Unity — Periodic Sum
nta_pyq_2026_jan
Grade 11
Question:
If $x^2+x+1=0$, then the value of $\left(x+\dfrac{1}{x}\right)^4+\left(x^2+\dfrac{1}{x^2}\right)^4+\left(x^3+\dfrac{1}{x^3}\right)^4+\cdots+\left(x^{25}+\dfrac{1}{x^{25}}\right)^4$ is:
Step-by-Step Solution
Key Concept: $x=\omega$ (primitive cube root of unity), $x^3=1$. $x+1/x=-1$, $x^2+1/x^2=(-1)^2-2=x^{2n}+x^{-2n}$. Period 3: if $n\equiv0\pmod3$: $x^n+x^{-n}=2$; if $n\equiv1,2\pmod3$: $x^n+x^{-n}=-1$.
8 multiples of 3 give 128; 17 others give 17. Total $=145$.
Correct Answer: 3