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Mathematics Question on Complex Numbers and Quadratic Equations

If α,β,γ\alpha ,\beta,\gamma are the roots of the equation x3+4x+2=0x^3+4x+2=0 then α3+β3+γ3\alpha^3+\beta^3+\gamma^3

A

2

B

6

C

-2

D

-6

Answer

-6

Explanation

Solution

The correct answer is D:-6
Given that;
For equation: x3+4x+2=0;x^3+4x+2=0; α,β,γ\alpha,\beta,\gamma are there roots
∴α+β+γ=0\therefore \alpha+\beta+\gamma=0 (coefficient  of  x2Coefficient  of  x3)\frac{coefficient\space of\space x^2}{Coefficient\space of\space x^3})
αβ+βγ+αγ=4\alpha\beta+\beta\gamma+\alpha\gamma=4 (coefficient  of  xcoefficient  of  x3)(\frac{coefficient\space of\space x}{coefficient\space of\space x^3})
αβγ=−2\alpha\beta\gamma=-2, (constantcoefficient  of  x3=−2)(\frac{constant}{coefficient\space of\space x^3}=-2)
we know that,
a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
∴α3+β3+γ3−3αβγ=(α+β+γ)(α2+β2+γ2−αβ−βγ−γα)\therefore \alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma=(\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)
⇒α3+β3+γ3=−6\alpha^3+\beta^3+\gamma^3=-6
alpha