Let $α, β, γ, δ$ are distinct imaginary roots of $z^{5} = 1,$ then value of $|\begin{array}{lll} e^{α} e^{2 α} e^{3 α + 1} - e^{- δ} \\ e^{β} e^{2 β} e^{3 β + 1} - e^{- δ} \\ e^{γ} e^{2 γ} e^{3 γ + 1} - e^{- δ} \end{array}|$ is

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$e^{5}$

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Detailed Solution

$\begin{array}{l} |\begin{array}{ccc} e^{α} & e^{2 α} & e^{3 α + 1} \\ e^{β} & e^{2 β} & e^{3 β + 1} \\ e^{γ} & e^{2 γ} & e^{3 γ + 1} \end{array}| - |\begin{array}{ccc} e^{α} & e^{2 α} & e^{- δ} \\ e^{β} & e^{2 β} & e^{- δ} \\ e^{γ} & e^{2 γ} & e^{- δ} \end{array}| \\ = e^{α} \cdot e^{β} \cdot e^{γ} \cdot e^{1} |\begin{array}{ccc} 1 & e^{α} & e^{2 α} \\ 1 & e^{β} & e^{2 β} \\ 1 & e^{γ} & e^{2 γ} \end{array}| - e^{- δ} |\begin{array}{lll} 1 & e^{α} & e^{2 α} \\ 1 & e^{β} & e^{2 β} \\ 1 & e^{γ} & e^{2 γ} \end{array}| \\ = (e^{α + β + γ + 1} - e^{- δ}) |\begin{array}{lll} 1 & e^{α} & e^{2 α} \\ 1 & e^{β} & e^{2 β} \\ 1 & e^{γ} & e^{2 γ} \end{array}| = 0 ∴ α + β + γ + δ + 1 = 0 \end{array}$