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Q.

If $α, β$ are the roots of $x^{2} + p x + q = 0, ω^{3} = 1$ then $(ω α + ω^{2} β) \cdot (ω^{2} α + ω β)$ is

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a

$p^{2} - q$

b

$p^{2} + q$

c

$p^{2} - 3 q$

d

$p^{2} + 3 q$

answer is C.

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Detailed Solution

$\begin{array}{l} α + β = - p, α β = q \\ (ω α + ω^{2} β) (ω^{2} α + ω β) \\ = α^{2} + β^{2} + (ω^{2} + ω) α β = α^{2} + β^{2} - α β \\ (∴ ω^{3} = 1, 1 + ω + ω^{2} = 0) \\ = (α + β)^{2} - 3 α β = p^{2} - 3 q \end{array}$

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