If alpha, beta are the roots of ax^2+bx+c=0,t | vedclass

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(A) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2+bx+c=0$. Since $\alpha$ is a root,$a\alpha^2+b\alpha+c=0$,which impl

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(A) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2+bx+c=0$. Since $\alpha$ is a root,$a\alpha^2+b\alpha+c=0$,which implies $a\alpha^2+b\alpha = -c$. Factoring out $\alpha$,we get $\alpha(a\alpha+b) = -c$,so $a\alpha+b = -\frac{c}{\alpha}$. Similarly,since $\beta$ is a root,$a\beta^2+b\beta+c=0$,which implies $a\beta^2+b\beta = -c$. Factoring out $\beta$,we get $\beta(a\beta+b) = -c$,so $a\beta+b = -\frac{c}{\beta}$. Now,substitute these into the given expression: $\left(\frac{\alpha}{a\beta+b}\right)^3 - \left(\frac{\beta}{a\alpha+b}\right)^3 = \left(\frac{\alpha}{-c/\beta}\right)^3 - \left(\frac{\beta}{-c/\alpha}\right)^3$ $= \left(-\frac{\alpha\beta}{c}\right)^3 - \left(-\frac{\alpha\beta}{c}\right)^3$ $= -\left(\frac{\alpha\beta}{c}\right)^3 + \left(\frac{\alpha\beta}{c}\right)^3 = 0$.

If $\alpha, \beta$ are the roots of $ax^2+bx+c=0$,then $\left(\frac{\alpha}{a\beta+b}\right)^3 - \left(\frac{\beta}{a\alpha+b}\right)^3 = $

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