A) \[1\]
B) \[-1\]
C) \[2\]
D) \[0\]
Correct Answer: B
Solution :
Given, equation is \[{{x}^{2}}+x+1=0\] \[\Rightarrow \] \[x=\frac{-1\pm \sqrt{3}i}{2}\] \[\Rightarrow \] \[x=\omega ,\,{{\omega }^{2}}\] Since, \[\alpha \] and \[\beta \] are the roots of \[{{x}^{2}}+x+1=0\] \[\therefore \] \[\alpha =\omega \] and \[\beta ={{\omega }^{2}}\] Now, \[{{\alpha }^{16}}+{{\beta }^{16}}={{(\omega )}^{16}}+{{({{\omega }^{2}})}^{16}}\] \[={{\omega }^{16}}+{{\omega }^{22}}\] \[=\omega +{{\omega }^{2}}\] \[(\because \,{{\omega }^{3}}=1)\] \[=-1\] \[(\because \,\,1+\omega +{{\omega }^{2}}=0)\]
You need to login to perform this action.
You will be redirected in
3 sec
