If α, β are roots of the equation x2+2x+4=0, then 1α3+1β3 is equal to

Here, α+β=-2 and αβ=4∴1α3+1β3=α3+β3(αβ)3=(α+β)3-3αβ(α+β)(αβ)3=(-2)3-3(-2)(4)(4)3=1664=14

If α, β are roots of the equation x2+2x+4=0, then 1α3+1β3 is equal to

Here, α+β=-2 and αβ=4∴1α3+1β3=α3+β3(αβ)3=(α+β)3-3αβ(α+β)(αβ)3=(-2)3-3(-2)(4)(4)3=1664=14

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If $α, β$ are roots of the equation $x^{2} + 2 x + 4 = 0$ , then $\frac{1}{α^{3}} + \frac{1}{β^{3}}$ is equal to

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$- \frac{1}{2}$

$\frac{1}{2}$

$32$

$\frac{1}{4}$

answer is D.

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Here, $α + β = - 2 and αβ = 4$

$∴ \frac{1}{α^{3}} + \frac{1}{β^{3}} = \frac{α^{3} + β^{3}}{{(αβ)}^{3}} = \frac{{(α + β)}^{3} - 3 αβ (α + β)}{{(αβ)}^{3}}$

$= \frac{{(- 2)}^{3} - 3 (- 2) (4)}{{(4)}^{3}} = \frac{16}{64} = \frac{1}{4}$

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If α, β are roots of the equation x2+2x+4=0, then 1α3+1β3 is equal to

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If $α, β$ are roots of the equation $x^{2} + 2 x + 4 = 0$ , then $\frac{1}{α^{3}} + \frac{1}{β^{3}}$ is equal to