Let p and q be real numbers such that p neq 0 | vedclass

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(B) Given $\alpha+\beta = -p$ and $\alpha^3+\beta^3 = q$. Using the identity $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$,we hav

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$(p^3+q)x^2-(p^3-2q)x+(p^3+q)=0$

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(B) Given $\alpha+\beta = -p$ and $\alpha^3+\beta^3 = q$. Using the identity $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$,we have $q = (-p)^3 - 3\alpha\beta(-p) = -p^3 + 3p\alpha\beta$. Thus,$3p\alpha\beta = p^3+q$,which implies $\alpha\beta = \frac{p^3+q}{3p}$. The quadratic equation with roots $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is $x^2 - (\frac{\alpha}{\beta} + \frac{\beta}{\alpha})x + 1 = 0$. This simplifies to $x^2 - (\frac{\alpha^2+\beta^2}{\alpha\beta})x + 1 = 0$. Since $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (-p)^2 - 2(\frac{p^3+q}{3p}) = p^2 - \frac{2(p^3+q)}{3p} = \frac{3p^3 - 2p^3 - 2q}{3p} = \frac{p^3-2q}{3p}$. Substituting these into the equation: $x^2 - (\frac{(p^3-2q)/3p}{(p^3+q)/3p})x + 1 = 0$. $x^2 - (\frac{p^3-2q}{p^3+q})x + 1 = 0$. Multiplying by $(p^3+q)$,we get $(p^3+q)x^2 - (p^3-2q)x + (p^3+q) = 0$.

Let $p$ and $q$ be real numbers such that $p \neq 0$,$p^3 \neq q$,and $p^3 \neq -q$. If $\alpha$ and $\beta$ are nonzero complex numbers satisfying $\alpha+\beta = -p$ and $\alpha^3+\beta^3 = q$,then a quadratic equation having $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ as its roots is

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