JEE Main 2025 — Mathematics Algebra

$$x^2+x+1=0$$x2+x+1=0

$\alpha$α is root

$\begin{aligned}
& \therefore \alpha^2+\alpha+1=0 \
& \Rightarrow \alpha=\omega \text { as } \omega^2 \text { [cube root of unity] }
\end{aligned}$

also

$$\left.\begin{array}{l}<br/>{\left[\begin{array}{ll}<br/>4-a-2 b & 64-a-14 b<br/>\end{array} 52+2 a-8 b\right.}<br/>\end{array}\right]$$<br/>[<br/>4−a−2b​64−a−14b<br/>​52+2a−8b<br/>​]

$$\begin{aligned} & =\left[\begin{array}{ll}0 & 0&0\end{array}\right] \\ \therefore & a+2 b=4 \\ & a+14 b=64 \\ \Rightarrow & 12 b=60 \Rightarrow b=5 \\ \Rightarrow & a=-6 \\ \therefore & \frac{4}{\alpha^4}+\frac{m}{\alpha^{-6}}+\frac{n}{\alpha^5}=3 \\ \Rightarrow & \frac{4}{\omega}+\frac{m}{1}+\frac{n}{\omega^2}=3 \\ \Rightarrow & 4 \omega^2+m+n \omega=3\end{aligned}$$∴⇒⇒∴⇒⇒​=[0​0​0​]a+2b=4a+14b=6412b=60⇒b=5a=−6α44​+α−6m​+α5n​=3ω4​+1m​+ω2n​=34ω2+m+nω=3​

$\begin{aligned}
& \Rightarrow 4\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)+\mathrm{m}+\mathrm{n}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)=3 \
& \therefore-2+\mathrm{m}-\frac{\mathrm{n}}{2}=3....(1) \
& & \frac{-4 \sqrt{3}}{2}+\frac{\mathrm{n} \sqrt{3}}{2}=0 \
& \therefore \mathrm{n}=4 \
& \mathrm{m}=7 \
& \therefore \mathrm{m}+\mathrm{n}=11
\end{aligned}$