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TS EAMCET · Maths · Quadratic Equation

Let \(\alpha, \beta, \gamma\) be the roots of \(x^3+x+10=0\) and \(\alpha_1=\frac{\alpha+\beta}{\gamma^2}, \beta_1=\frac{\beta+\gamma}{\alpha^2}, \gamma_1=\frac{\gamma+\alpha}{\beta^2}\). Then, the value of \(\left(\alpha_1^3+\beta_1^3+\gamma_1^3\right)-\frac{1}{10}\left(\alpha_1^2+\beta_1^2+\gamma_1^2\right)\) is

  1. A \(\frac{1}{10}\)
  2. B \(\frac{1}{5}\)
  3. C \(\frac{3}{10}\)
  4. D \(\frac{1}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{3}{10}\)

Step-by-step Solution

Detailed explanation

Since, \(\alpha, \beta, \gamma\) are the roots of the equation \[ \begin{aligned} & x^3+x+10=0 . \\ & \therefore \quad \alpha+\beta+\gamma=0 \end{aligned} \] Now, \(\alpha_1=\frac{\alpha+\beta}{\gamma^2}=\frac{-\gamma}{\gamma^2}=\frac{-1}{\gamma}\) Similarly,…
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