Question Number 140746 by 676597498 last updated on 12/May/21
Answered by Ar Brandon last updated on 12/May/21
$$\left(\mathrm{i}\right)\:\left(\mathrm{1}+\mathrm{3w}\right)\left(\mathrm{1}+\mathrm{3w}^{\mathrm{2}} \right)=\mathrm{1}+\mathrm{3}\left(\mathrm{w}^{\mathrm{2}} +\mathrm{w}\right)+\mathrm{9w}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\mathrm{3}\left(−\mathrm{w}^{\mathrm{3}} \right)+\mathrm{9}\left(\mathrm{1}\right)=\mathrm{1}−\mathrm{3}+\mathrm{9}=\mathrm{7} \\ $$$$ \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{1}+\mathrm{3w}+\mathrm{w}^{\mathrm{2}} =\mathrm{1}+\mathrm{3w}+\left(−\mathrm{w}−\mathrm{w}^{\mathrm{3}} \right)=\mathrm{1}+\mathrm{2w}−\mathrm{1}=\mathrm{2w} \\ $$
Commented by Ar Brandon last updated on 12/May/21
$$\mathrm{1}+\mathrm{w}+\mathrm{w}^{\mathrm{2}} =\mathrm{0}\:\left(\mathrm{w}^{\mathrm{3}} =\mathrm{1}\right) \\ $$
Answered by peter frank last updated on 12/May/21
$$\left({i}\right)\mathrm{1}+\mathrm{3}{w}^{\mathrm{2}} +\mathrm{3}{w}+\mathrm{9}{w}^{\mathrm{3}} \\ $$$${w}^{\mathrm{3}} =\mathrm{1}\:\:\mathrm{1}+{w}+{w}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{3}\left(−\mathrm{1}\right)+\mathrm{9}\left(\mathrm{1}\right)=\mathrm{1}−\mathrm{3}+\mathrm{9}=\mathrm{7} \\ $$$$\left.{ii}\right)\mathrm{1}+\mathrm{3}\left(−\mathrm{1}\right)=−\mathrm{2} \\ $$