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Question Number 191611 by vishal1234 last updated on 27/Apr/23

if x^2 −x+1 = 0 and α and β are thd roots of this equation then evaluate ((α^(100) +β^(100) )/(α^(100) −β^(100) ))

$${if}\:{x}^{\mathrm{2}} −{x}+\mathrm{1}\:=\:\mathrm{0}\:{and}\:\alpha\:{and}\:\beta\:{are}\:{thd}\:{roots} \\ $$$${of}\:{this}\:{equation}\:{then}\:{evaluate}\:\frac{\alpha^{\mathrm{100}} +\beta^{\mathrm{100}} }{\alpha^{\mathrm{100}} −\beta^{\mathrm{100}} } \\ $$

Answered by mehdee42 last updated on 27/Apr/23

x=((1±(√3)i)/2)⇒α=((1+(√3)i)/2)=e^((π/3)i) & β=((1−(√3)i)/2)=e^(−(π/3)i) α^(100) +β^(100) =e^(((100π)/3)i) +e^(−((100π)/3)i) =2cos((100π)/3)=−2cos(π/3)=−1 α^(100) −β^(100) =e^(((100π)/3)i) +e^(−((100π)/3)i) =2sin((100π)/3)=−2isin(π/3)=−(√3)i ⇒answer=(1/( (√3)i))=−((√3)/3)i

$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}{i}}{\mathrm{2}}\Rightarrow\alpha=\frac{\mathrm{1}+\sqrt{\mathrm{3}}{i}}{\mathrm{2}}={e}^{\frac{\pi}{\mathrm{3}}{i}} \:\&\:\beta=\frac{\mathrm{1}−\sqrt{\mathrm{3}}{i}}{\mathrm{2}}={e}^{−\frac{\pi}{\mathrm{3}}{i}} \\ $$$$\alpha^{\mathrm{100}} +\beta^{\mathrm{100}} ={e}^{\frac{\mathrm{100}\pi}{\mathrm{3}}{i}} +{e}^{−\frac{\mathrm{100}\pi}{\mathrm{3}}{i}} =\mathrm{2}{cos}\frac{\mathrm{100}\pi}{\mathrm{3}}=−\mathrm{2}{cos}\frac{\pi}{\mathrm{3}}=−\mathrm{1} \\ $$$$\alpha^{\mathrm{100}} −\beta^{\mathrm{100}} ={e}^{\frac{\mathrm{100}\pi}{\mathrm{3}}{i}} +{e}^{−\frac{\mathrm{100}\pi}{\mathrm{3}}{i}} =\mathrm{2}{sin}\frac{\mathrm{100}\pi}{\mathrm{3}}=−\mathrm{2}{isin}\frac{\pi}{\mathrm{3}}=−\sqrt{\mathrm{3}}{i} \\ $$$$\Rightarrow{answer}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}{i}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{i} \\ $$$$ \\ $$

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