If $f(\alpha) = \left| \begin{matrix} 1 & \alpha & \alpha^{2} \ \alpha & \alpha^{2} & 1 \ \alpha^{2} & 1 & \alpha \end{matrix} ight|$, then $f(\sqrt[3]{3})$ is equal to

Question

If $f(\alpha) = \left| \begin{matrix} 1 & \alpha & \alpha^{2} \ \alpha & \alpha^{2} & 1 \ \alpha^{2} & 1 & \alpha \end{matrix} ight|$, then $f(\sqrt[3]{3})$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Given, $f(\alpha) = \left| \begin{matrix}
1 & \alpha & \alpha^{2} \
\alpha & \alpha^{2} & 1 \
\alpha^{2} & 1 & \alpha
\end{matrix} ight|$
$= 1\left( \alpha^{3} - 1 ight) - \alpha\left( \alpha^{2} - \alpha^{2} ight) + \alpha^{2}(\alpha - \alpha^{4})$
$= \alpha^{3} - 1 - 0 + \alpha^{3} - \alpha^{6}$
$\Rightarrow f(\sqrt[3]{3} = 3 - 1 - 0 + 3 - 3^{2} = 6 - 10 = - 4$

Q. If $f(\alpha) = \left| \begin{matrix} 1 & \alpha & \alpha^{2} \\ \alpha & \alpha^{2} & 1 \\ \alpha^{2} & 1 & \alpha \end{matrix} \right|$, then $f(\sqrt[3]{3})$ is equal to

Explanation

Reviewed by Dr. Abubacker

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