TS EAMCET · Maths · Quadratic Equation
\(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+2 x^2-x-2=0\) then \(\alpha^6+\beta^6+\gamma^6=\)
- A \(3\)
- B \(129\)
- C \(68\)
- D \(192\)
Answer & Solution
Correct Answer
(C) \(68\)
Step-by-step Solution
Detailed explanation
\(\alpha+\beta+\gamma=-2 ; \alpha \beta+\beta \gamma+\gamma \alpha=-1 ; \alpha \beta \gamma=2\) \(\begin{aligned} & x^3+2 x^2-x-2=0 \\ & \Rightarrow(x+1)(x-1)(x+2)=0 \\ & \therefore x=1,-1,-2 \\ & \alpha^6+\beta^6+\gamma^6=(1)^6+(-1)^6+(-2)^6=66\end{aligned}\)
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