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(D) Given the cubic equation $x^3+px^2+qx+r=0$,the roots are $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -p$ $\alpha\beta+\be

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$pq-r$

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(D) Given the cubic equation $x^3+px^2+qx+r=0$,the roots are $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -p$ $\alpha\beta+\beta\gamma+\gamma\alpha = q$ $\alpha\beta\gamma = -r$ We know that $\alpha+\beta = -p-\gamma$,$\beta+\gamma = -p-\alpha$,and $\gamma+\alpha = -p-\beta$. Therefore,$(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = (-p-\gamma)(-p-\alpha)(-p-\beta) = -(p+\gamma)(p+\alpha)(p+\beta)$. Let $f(x) = (x-\alpha)(x-\beta)(x-\gamma) = x^3+px^2+qx+r$. Then $f(-p) = (-p-\alpha)(-p-\beta)(-p-\gamma) = (-p)^3+p(-p)^2+q(-p)+r = -p^3+p^3-pq+r = r-pq$. Since $f(-p) = -(p+\alpha)(p+\beta)(p+\gamma)$,we have $-(p+\alpha)(p+\beta)(p+\gamma) = r-pq$. Thus,$(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = pq-r$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+px^2+qx+r=0$,then $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=$

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