Let alpha be a root of x^2+x+1=0 and suppose | vedclass

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(D) The roots of $x^2+x+1=0$ are $\omega$ and $\omega^2$,where $\omega$ is a complex cube root of unity.We know that $1+\omega+\omega^2=0$.For $\alpha

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$\frac{2}{9}$

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(D) The roots of $x^2+x+1=0$ are $\omega$ and $\omega^2$,where $\omega$ is a complex cube root of unity.We know that $1+\omega+\omega^2=0$.For $\alpha^a+\alpha^b+\alpha^c=0$,the powers $\alpha^a, \alpha^b, \alpha^c$ must be a permutation of ${1, \omega, \omega^2}$.This implies that $a, b, c$ must be of the form $3k_1+r_1, 3k_2+r_2, 3k_3+r_3$ where ${r_1, r_2, r_3} = {0, 1, 2}$ modulo $3$.In a die,the numbers are ${1, 2, 3, 4, 5, 6}$.Modulo $3$,these are ${1, 2, 0, 1, 2, 0}$.There are two $1$s,two $2$s,and two $0$s.To have ${r_1, r_2, r_3} = {0, 1, 2}$,we need to choose one number from each set of residues.Number of ways to choose $a, b, c$ such that their residues are ${0, 1, 2}$ in any order is $3! 	imes (2 	imes 2 	imes 2) = 6 	imes 8 = 48$.Total outcomes $= 6^3 = 216$.Probability $= \frac{48}{216} = \frac{2}{9}$.

List-$I$	List-$II$
$A. P(E_2) =$	$I. 2/3$
$B. P(E_1 \| E_2) =$	$II. 5/6$
$C. P(\bar{E}_2 \| E_1) =$	$III. 1/3$
$D. P(\bar{E}_1 \cup \bar{E}_2) =$	$IV. 1/2$

Let $\alpha$ be a root of $x^2+x+1=0$ and suppose that a fair die is thrown $3$ times. If $a, b,$ and $c$ are the numbers shown on the die,then the probability that $\alpha^a+\alpha^b+\alpha^c=0$ is

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