COMEDK · Maths · 2. Quadratic Equation
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^{3}-3 x^{2}+2 x-1=0\), then the value of \((1-\alpha)(1-\beta)(1-\gamma)\) is
- A 1
- B 2
- C \(-1\)
- D \(-2\)
Answer & Solution
Correct Answer
(C) \(-1\)
Step-by-step Solution
Detailed explanation
We have, \(x^{3}-3 x^{2}+2 x-1\) So, \(\quad \alpha+\beta+\gamma=3\) \(\alpha \beta+\beta \gamma+\gamma \alpha=2\) \(\alpha \beta \gamma=1\) Now, \((1-\alpha)(1-\beta)(1-\gamma)\) \(=1-(\alpha+\beta+\gamma)+(\alpha \beta+\beta \gamma+\gamma \alpha)-\alpha \beta \gamma\)…
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