Question

If $\alpha ,\beta,\gamma$ are the roots of the equation $x^3+4x+2=0$ then $\alpha^3+\beta^3+\gamma^3$

• A. 2
• B. 6
• C. -2
• D. -6

Answer 1

The Correct Option is D

The correct answer is D:-6
Given that;
For equation: \(x^3+4x+2=0;\)  \(\alpha,\beta,\gamma\) are there roots
\(\therefore \alpha+\beta+\gamma=0\)  (\(\frac{coefficient\space of\space x^2}{Coefficient\space of\space x^3})\)
\(\alpha\beta+\beta\gamma+\alpha\gamma=4\)  \((\frac{coefficient\space of\space x}{coefficient\space of\space x^3})\)
\(\alpha\beta\gamma=-2\),  \((\frac{constant}{coefficient\space of\space x^3}=-2)\)
we know that,
\(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\)
\(\therefore \alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma=(\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)\)
⇒\(\alpha^3+\beta^3+\gamma^3=-6\)