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Can someone explain to me why: If \omega,\omega_2,...\omega_n are the nth roots of unity, then \sum_{i=1}^{n} \omega _i=0

(a)\,\,(a+b+c)^3\\=a^3+b^3+c^3+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)+6abc\\=a^3+b^3+c^3+3(ab(a+b+c)+bc(a+b+c)+ac(a+b+c))-3abc\\=a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)-3abc\\\\\therefore a^3+b^3+c^3-3abc\\=(a+b+c)^3-3(a+b+c)(ab+bc+ac)\\=(a+b+c)((a+b+c)^2-3(ab+bc+ac))\\=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)...

(a)\,\,S_n=\sum_{j=0}^{n}\int_{0}^{x}\cos[(2j+1)y]dy\\=\sum_{j=0}^{n}\frac{\sin[(2j+1)x]}{2j+1} (b)(i)\,\, 2\cos\theta=cis\theta+cis(-\theta)\\2\cos\theta=e^{i\theta}+e^{-i\theta}\\\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})\\\cos[(2j+1)y]=\frac{1}{2}(e^{iy(2j+1)}+e^{-iy(2j+1)})$by...