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Using the root-coefficient relationship, we have the following: $$\alpha + \beta + \gamma = -1,$$ $$\alpha\beta\gamma = -1,$$ $$\alpha\beta + \beta\gamma + \gamma ...
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(2) $$\frac{\alpha^7+\beta^7+\gamma^7}{7} = \left(\frac{\alpha^5+\beta^5+\gamma^5}{5} \right)\left(\frac{\alpha^2+\beta^2+\gamma^2}{2}\right)$$ I found this interesting problem from a teacher and it got stuck in my head, so if possible I request some hints to solving this problem.
Also $\alpha\beta+\beta\gamma+\gamma\alpha=-\alpha\beta\gamma$ so, yes, I can see why you are asking...$\alpha^3+\beta^3+\gamma^3=3\alpha\beta\gamma$ $\endgroup$ – Martin Hansen Commented Jul 11, 2020 at 15:30
Hints: Note that $\alpha + \beta + \gamma = 0$ ; $\alpha\beta + \alpha\gamma + \beta\gamma = -p $ and $\alpha\beta\gamma = -q $.
Find the minimum of $\\sin \\alpha+\\sin \\beta+\\sin \\gamma,$ where $\\alpha,\\beta,\\gamma\\in \\mathbb{R}$ satisfying $\\alpha+\\beta+\\gamma = \\pi$. Options: (a ...
Based on your expressions for the first and second raw moments, I will assume that the gamma distribution is parametrized by shape α and scale β; i.e., fY(y) = yα−1e−y/β βαΓ(α), y> 0. In such a case, equating on raw (uncentered) sample moments gives the system. y¯1 y¯2 = αβ, =α2β2 + αβ2 = α(α + 1)β2.
Using Vieta's formulas, $$\alpha+\beta+\gamma=\frac{-q}{p}$$ $$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{r}{p}$$ $$\alpha\beta\gamma=\frac{-s}{p}$$ I first thought that this would help. We can find $\alpha\beta\gamma$ by Vieta's formula but the expression inside this bracket could not be possibly found.
If $\\alpha, \\beta,\\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that: $$(\\frac{1}{\\beta^3}+\\frac{1}{\\gamma^3}-\\frac{1}{\\alpha^3})(\\frac{1}{\\gamma^3 ...
It’s not clear why you have written $$\frac{\gamma^2-2\alpha\beta}{\gamma^2}=\frac{1-2\alpha\beta\gamma}{\gamma^2\gamma}$$ This would mean that $\gamma^3=1$, which is not true. It should be obvious to you that your answer is incorrect as you have arrived at a quadratic polynomial, not a cubic.