AP EAMCET · Maths · Complex Number
Let \(\alpha\) be a root of \(x^2+x+\mathrm{I}=0\) 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 & Solution
Correct Answer
(D) \(\frac{2}{9}\)
Step-by-step Solution
Detailed explanation
Total numbers of ways for \((a, b, c)=6 \times 6 \times 6\) Here, \(\begin{aligned} & \omega, \omega^2 \\ & \omega^3=1, \\ & \omega^4=\omega, \\ & \omega^5=\omega^2, \end{aligned}\) and \(\omega^6=1\) Since, \(\omega^2\), \(\omega\) are roots of \(x^2+x+1=0\).…
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