JEE Advanced 2026 · Paper 1, Mathematics question 13

Let $\omega$ be a complex cube root of unity, so $1+\omega+\omega^2=0$.

(P) $\alpha=\omega$, $\beta=\omega^2$; $\alpha+1=-\omega^2$, $\beta+1=-\omega$. With $2026\equiv1\pmod3$:
$(-\omega^2)^{2026}=\omega^{2}$ and $(-\omega)^{2026}=\omega$. The roots are $\omega^{-2}=\omega$ and $\omega^{-1}=\omega^2$: equation $x^2+x+1=0$ → 1.

(Q) The exponent is odd: $(-\omega^2)^{2027}=-\omega$, $(-\omega)^{2027}=-\omega^2$. The roots are $-\omega^2,-\omega$, with sum $1$ and product $1$: $x^2-x+1=0$ → 2.

(R) $\gamma=-\omega$, $\delta=-\omega^2$, so $\gamma-1=\omega^2$, $\delta-1=\omega$. The sum is $\omega^{-4052}+\omega^{-2026}=\omega+\omega^2=-1$ → 4.

(S) $u=p+1$ satisfies $u^2-u-1=0$, so $v=1/u$ satisfies $v^2+v-1=0$: $v_1+v_2=-1$, $v_1v_2=-1$.
$$v_1^3+v_2^3=(-1)^3-3(-1)(-1)=-4$$ → 5.

P→1, Q→2, R→4, S→5.