Question

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

• A. $\frac{2}{36}$
• B. $\frac{1}{27}$
• C. $\frac{1}{72}$
• D. $\frac{2}{9}$

Answer 1

Answer: (D)

Total numbers of ways for $(a, b, c)=6 \times 6 \times 6$
Here, $\omega, \omega^{2}$
$\omega^{3} =1 $
$\omega^{4}=\omega $
$\omega^{5} =\omega^{2} $
and $\omega^{6} =1$
Since, $\omega^{2}, \omega$ are roots of $x^{2}+x+1=0$
$\because \omega^{a}+\omega^{b}+\omega^{c}=0$
Now, suitable values for $(a, b, c)=6 \times 4 \times 2$
$\therefore $ Required probability $=\frac{6 \times 4 \times 2}{6 \times 6 \times 6}=\frac{8}{36}=\frac{2}{9}$