Let $\alpha$ be a solution of $x^2+x+1=0$, and for some $a$ and $b$ in
$\mathbb{R},\left[\begin{array}{lll}4 & \mathrm{a} & \mathrm{b}\end{array}\right]\left[\begin{array}{ccc}1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8\end{array}\right]=\left[\begin{array}{ccc}0 & 0 & 0\end{array}\right]$. If $\frac{4}{\alpha^4}$ $+\frac{\mathrm{m}}{\alpha^{\mathrm{a}}}+\frac{\mathrm{n}}{\alpha^{\mathrm{b}}}=3$, then $\mathrm{m}+\mathrm{n}$ is equal to

Question

Aicharya · Accepted Answer

$x^2+x+1=0$
$\alpha$ 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
$\left.\begin{array}{l}
{\left[\begin{array}{ll}
4-a-2 b & 64-a-14 b
\end{array} 52+2 a-8 b\right.}
\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{ll}0 & 0&0\end{array}\right] \\ \therefore & a+2 b=4 \\ & a+14 b=64 \\ \Rightarrow & 12 b=60 \Rightarrow b=5 \\ \Rightarrow & a=-6 \\ \therefore & \frac{4}{\alpha^4}+\frac{m}{\alpha^{-6}}+\frac{n}{\alpha^5}=3 \\ \Rightarrow & \frac{4}{\omega}+\frac{m}{1}+\frac{n}{\omega^2}=3 \\ \Rightarrow & 4 \omega^2+m+n \omega=3\end{aligned}$
$\begin{aligned}
& \Rightarrow 4\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)+\mathrm{m}+\mathrm{n}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)=3 \\ & \therefore-2+\mathrm{m}-\frac{\mathrm{n}}{2}=3....(1) \\ & \& \frac{-4 \sqrt{3}}{2}+\frac{\mathrm{n} \sqrt{3}}{2}=0 \\ & \therefore \mathrm{n}=4 \\ & \mathrm{~m}=7 \\ & \therefore \mathrm{~m}+\mathrm{n}=11
\end{aligned}$

Let $\alpha$ be a solution of $x^2+x+1=0$, and for some $a$ and $b$ in $\mathbb{R},\left[\begin{array}{lll}4…

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