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Mathematics Question on Complex Numbers and Quadratic Equations

If α\alpha and β\beta are the roots of the equation x2+2x+4=0x^{2} + 2x + 4 = 0, then 1α3+1β3\frac{1}{\alpha^{3}}+\frac{1}{\beta^{3}} is equal to

A

−1/2-1/2

B

1/21/2

C

3232

D

1/41/4

Answer

1/41/4

Explanation

Solution

Given equation is x2+2x+4=0x^{2} + 2x +4 = 0 Since α,β\alpha, \beta are roots of this equation. ∴ α+β=−2\therefore\, \alpha+\beta=-2 and αβ=4\alpha\beta=4 Now, 1α3+1β3\frac{1}{\alpha^{3}}+\frac{1}{\beta^{3}} =α3+β3(αβ)3=\frac{\alpha^{3}+\beta^{3}}{\left(\alpha\beta\right)^{3}} =(α+β)(α2+β2−αβ)(αβ)3=\frac{\left(\alpha+\beta\right)\left(\alpha^{2}+\beta^{2}-\alpha\beta\right)}{\left(\alpha\beta\right)^{3}} =(−2)((α+β)2−3αβ)4×4×4=\frac{\left(-2\right)\left(\left(\alpha+\beta\right)^{2}-3\alpha\beta\right)}{4\times4\times4} =−2(4−12)4×4×4=\frac{-2\left(4-12\right)}{4\times4\times4} =(−2)×(−8)4×4×4=14=\frac{\left(-2\right)\times\left(-8\right)}{4\times4\times4}=\frac{1}{4}