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(D) Given the cubic equation $x^{3}+4x+2=0$. Let the roots be $\alpha, \beta, \gamma$. From the properties of roots of a cubic equation $ax^{3}+bx^{2}

Vedclass · Accepted Answer

$-6$

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(D) Given the cubic equation $x^{3}+4x+2=0$. Let the roots be $\alpha, \beta, \gamma$. From the properties of roots of a cubic equation $ax^{3}+bx^{2}+cx+d=0$: Sum of roots $\Sigma \alpha = -\frac{b}{a} = 0$. Sum of roots taken two at a time $\Sigma \alpha \beta = \frac{c}{a} = 4$. Product of roots $\alpha \beta \gamma = -\frac{d}{a} = -2$. We know the identity $\alpha^{3}+\beta^{3}+\gamma^{3}-3 \alpha \beta \gamma = (\alpha+\beta+\gamma)(\alpha^{2}+\beta^{2}+\gamma^{2}-\alpha \beta-\beta \gamma-\gamma \alpha)$. Since $\Sigma \alpha = 0$,the right side becomes $0$. Therefore,$\alpha^{3}+\beta^{3}+\gamma^{3} = 3 \alpha \beta \gamma$. Substituting the value of the product of roots: $\alpha^{3}+\beta^{3}+\gamma^{3} = 3(-2) = -6$.

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^{3}+4x+2=0$,then $\alpha^{3}+\beta^{3}+\gamma^{3}$ is equal to

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