Question 1118051
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i)
{{{(w^2)^5 = w^10 = (w^5)^2 = 1^2 = 1}}}
{{{(w^3)^5 = w^15 = (w^5)^3 = 1^3 = 1}}}
{{{(w^4)^5 = w^20 = (w^5)^4 = 1^4 = 1}}}<br>
ii)
{{{1+w+w^2+w^3+w^4 = (1-w^5)/(1-w) = 0/(1-w) = 0}}}<br>
iii)
{{{w^(5n) = (w^5)^n = 1^n = 1}}}