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

A) 1
B) $- 4$
C) 4
D) 2
Explanation

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

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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