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Question Number 140746 by 676597498 last updated on 12/May/21

Answered by Ar Brandon last updated on 12/May/21

$$\left(\mathrm{i}\right)(1+3\mathrm{w})(1+{3\mathrm{w}}^2)=1+3({\mathrm{w}}^2+\mathrm{w})+{9\mathrm{w}}^3$$$$=1+3(-{\mathrm{w}}^3)+9\left(1\right)=1-3+9=7$$$$$$$$\left(\mathrm{ii}\right)1+3\mathrm{w}+{\mathrm{w}}^2=1+3\mathrm{w}+(-\mathrm{w}-{\mathrm{w}}^3)=1+2\mathrm{w}-1=2\mathrm{w}$$

Commented by Ar Brandon last updated on 12/May/21

$$1+\mathrm{w}+{\mathrm{w}}^2=0({\mathrm{w}}^3=1)$$

Answered by peter frank last updated on 12/May/21

$$\left(i\right)1+3w^2+3w+9w^3$$$$w^3=11+w+w^2=0$$$$1+3(-1)+9\left(1\right)=1-3+9=7$$$$ii)1+3(-1)=-2$$

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