11. Let $\alpha, \beta$ and $\gamma$ be the three roots of the equation $8 x^{3}+2012 x+2013=0$. Find the value of $(\alpha+\beta)^{3}+(\beta+\gamma)^{3}+(\gamma+\alpha)^{3}$.
(2 marks)
設 $\alpha$ 、 $\beta$ 和 $\gamma$ 為方程 $8 x^{3}+2012 x+2013=0$ 的三個根。求 $(\alpha+\beta)^{3}+(\beta+\gamma)^{3}+(\gamma+\alpha)^{3}$ 的值。
11. $\frac{6039}{8}$
11. Considering the sum of roots gives $\alpha+\beta+\gamma=0$. Hence
$$
(\alpha+\beta)^{3}+(\beta+\gamma)^{3}+(\gamma+\alpha)^{3}=(-\gamma)^{3}+(-\alpha)^{3}+(-\beta)^{3}=-\alpha^{3}-\beta^{3}-\gamma^{3} .
$$

As $\alpha$ is a root of the equation, we have $8 \alpha^{3}+2012 \alpha+2013=0$ and so $-\alpha^{3}=\frac{2012 \alpha+2013}{8}$. Likewise we have $-\beta^{3}=\frac{2012 \beta+2013}{8}$ and $-\gamma^{3}=\frac{2012 \gamma+2013}{8}$ and thus the answer is
$$
\frac{2012 \alpha+2013}{8}+\frac{2012 \beta+2013}{8}+\frac{2012 \gamma+2013}{8}=\frac{2012(\alpha+\beta+\gamma)+6039}{8}=\frac{6039}{8} .
$$