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Question Number 166372 by mathlove last updated on 19/Feb/22

Answered by MJS_new last updated on 19/Feb/22

$$\mathrm{Yes}\mathrm{I}\mathrm{can}.$$

Commented by mr W last updated on 19/Feb/22

$$yes,wecan!$$

Commented by peter frank last updated on 19/Feb/22

$$\mathrm{hahahhahah}$$

Answered by MathsFan last updated on 19/Feb/22

$$\mathbf{yes}\mathbf{i}\mathbf{can}!$$

Answered by Rasheed.Sindhi last updated on 19/Feb/22

$$b^2+b+1=0\Rightarrow (b-1)(b^2+b+1)=0$$$$\Rightarrow b^3-1=0\Rightarrow b=1,\omega ,{\omega }^2$$$$\because b=1istherootofb-1=0$$$$\therefore b^2+b+1\Rightarrow b=\omega ,{\omega }^2$$$$b=\omega :b^{2022}+\frac{1}{b^{2022}}={\omega }^{2022}+\frac{1}{{\omega }^{2022}}$$$$={\left({\omega }^3\right)}^{674}+\frac{1}{{\left({\omega }^3\right)}^{674}}=1^{674}+\frac{1}{1^{674}}=2$$$$b={\omega }^2:b^{2022}+\frac{1}{b^{2022}}={\left({\omega }^2\right)}^{2022}+\frac{1}{{\left({\omega }^2\right)}^{2022}}$$$$={\left({\omega }^3\right)}^{2\times 674}+\frac{1}{{\left({\omega }^3\right)}^{2\times 674}}=1^{2\times 674}+\frac{1}{1^{2\times 674}}=2$$$$Henceinbothcases:$$$$b^{2022}+\frac{1}{b^{2022}}=2$$

Commented by mathlove last updated on 19/Feb/22

$$thankssir$$

Commented by peter frank last updated on 19/Feb/22

$$\mathrm{thank}\mathrm{you}$$

Commented by Rasheed.Sindhi last updated on 19/Feb/22

$${you}^{\prime }rewelcome!$$

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