KCET · Maths · Quadratic Equation
If \(\alpha, \beta\) and \(\gamma\) are the roots of the equation \(x^{3}+4 x+2=0\), then \(\alpha^{3}+\beta^{3}+\gamma^{3}\) is equal to
- A 2
- B 6
- C \(-2\)
- D \(-6\)
Answer & Solution
Correct Answer
(D) \(-6\)
Step-by-step Solution
Detailed explanation
Given, \(x^{3}+4 x+2=0\)
\(\therefore \quad \Sigma \alpha=0, \quad \Sigma \alpha \beta=\frac{4}{1}=4, \quad \alpha \beta \gamma=\frac{-2}{1}=-2\)
\(\because \quad \Sigma \alpha=0\)
\(\therefore \quad \alpha^{3}+\beta^{3}+\gamma^{3}=3 \alpha \beta \gamma=3(-2)=-6\)
\(\therefore \quad \Sigma \alpha=0, \quad \Sigma \alpha \beta=\frac{4}{1}=4, \quad \alpha \beta \gamma=\frac{-2}{1}=-2\)
\(\because \quad \Sigma \alpha=0\)
\(\therefore \quad \alpha^{3}+\beta^{3}+\gamma^{3}=3 \alpha \beta \gamma=3(-2)=-6\)
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