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Question Number 104858 by ~blr237~ last updated on 24/Jul/20

$$(1-a^3)(1-b^3)......(1-x^3)(1-y^3)(1-z^3)????$$

Commented by 1549442205PVT last updated on 24/Jul/20

$$\mathbf{what}\mathbf{is}\mathbf{your}\mathbf{question}?$$

Answered by ~blr237~ last updated on 24/Jul/20

$$Justforlaughing:rememberweusuallyworkwej=e^{i\frac{2\pi }{3}}then{j}^3=1$$$$Andthatproductwillbenull$$

Commented by 1549442205PVT last updated on 25/Jul/20

$$\mathrm{j}={\mathrm{e}}^{\mathrm{i}\frac{2\pi }{3}}=\cos \frac{2\pi }{3}+\mathrm{isin}\frac{2\pi }{3}\Rightarrow {\mathrm{j}}^3={(\cos \frac{2\pi }{3}+\mathrm{isin}\frac{2\pi }{3})}^3$$$$=\mathbf{cos}2\mathit{\pi }+\mathbf{isin}2\mathit{\pi }=1,\mathbf{of}\mathbf{course},\mathbf{But}\mathbf{you}$$$${\mathbf{don}}^{\prime }\mathbf{t}\mathbf{give}\mathbf{this}\mathbf{hypothesis}\mathbf{in}\mathbf{your}$$$$\mathbf{question}.\mathbf{Nobody}\mathbf{can}\mathbf{guess}\mathbf{your}$$$$\mathbf{thought}$$

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