AP EAMCET · Maths · Quadratic Equation
If the sum of the cubes of the roots of the equation \(x^3-a x^2+b x-c=0\) is zero, then \(a^3+3 c=\)
- A \(-2 a b\)
- B \(2 \mathrm{ab}\)
- C \(-3 a b\)
- D \(3 a b\)
Answer & Solution
Correct Answer
(D) \(3 a b\)
Step-by-step Solution
Detailed explanation
Given: \(x^3-a x^2+b x-c=0 ...(i)\) Let \(\alpha, \beta \& \gamma\) are the roots equation (i), we get \(\begin{aligned} & \alpha+\beta+\gamma=a \\ & \alpha \beta+\beta \gamma+\gamma \alpha=b \\ & \alpha \beta \gamma=c \end{aligned}\) Also, given: \(\alpha^3+\beta^3+\gamma^3=0\)…
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