A) 32
B) 64
C) 128
D) 256
Correct Answer: C
Solution :
Given, \[\alpha ,\,\,\beta \]are the roots of\[{{x}^{2}}-2x+4=0\] \[\therefore \] \[\alpha +\beta =2\] ... (i) and \[\alpha \beta =4\] ...(ii) Now, \[\alpha -\beta =\sqrt{{{(\alpha +\beta )}^{2}}-4\alpha \beta }\] \[=\sqrt{4-4\times 4}=\sqrt{-12}\] \[\Rightarrow \] \[\alpha -\beta =2\sqrt{3}i\] ... (iii) On solving Eqs. (i) and (ii), we get \[\alpha =\frac{2+2\sqrt{3}i}{2}=-2\left( \frac{-1+\sqrt{3}i}{2} \right)=-2\omega \] And \[\beta =\frac{2-2\sqrt{3}i}{2}=-2\left( \frac{-1+\sqrt{3}i}{2} \right)=-2\omega \] Now, \[{{\alpha }^{6}}+{{\beta }^{6}}={{(-2{{\omega }^{2}})}^{6}}+{{(-2\omega )}^{6}}\] \[=64{{({{\omega }^{3}})}^{4}}+64{{({{\omega }^{3}})}^{2}}\] \[=128\] \[[\because \,\,{{\omega }^{3}}=1]\]
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