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Let $\mathrm{\alpha }$ be a root of equation $\mathrm{1+x^{2}+x^{4}=0 }$ . Then the value of $\mathrm{\alpha^{1011}+\alpha^{2022}-\alpha^{3033} }$ is equal to:
Option: 1
$1$

Option: 2
$\alpha$

Option: 3
$1+\alpha$

Option: 4
$1+2\alpha$

Answers (1)

best_answer

$\mathrm{1+x^{2}+x^{4}=0} \\$

$\mathrm{\text { Let } x^{2}=t} \\$

$\mathrm{\Rightarrow t^{2}+t+1=0 }\\$

$\mathrm{\Rightarrow t=\omega \text { or } \omega^{2}} \\$

$\mathrm{\Rightarrow x^{2}=\omega \text { or } \omega^{2}} \\$

$\mathrm{\Rightarrow \alpha^{2}=\omega \text { or } \omega^{2}}$

$\mathrm{\therefore \alpha^{1011}+\alpha^{2022}-\alpha^{3033}} \\$

$\mathrm{= \alpha\left(\alpha^{2}\right)^{505}+\left(\alpha^{2}\right)^{1011}-\alpha\left(\alpha^{2}\right)^{1516} \cdot \cdots(i)} \\$

$\mathrm{= \alpha \cdot \omega^{505}+\omega^{1011}-\alpha \cdot \omega^{1516}} \\$

$\mathrm{= \alpha \omega+1-\alpha \omega} \\$

$\mathrm{=1}$

We get the same answer by putting $\mathrm{\alpha^{2}=\omega^{2}}$ in (i)

Hence answer is option 1.

Posted by

himanshu.meshram

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