# x^3-x^2-x+3=0三根α,β,γ，求(α^2+α+1)(β^2+β+1)(γ^2+γ+1)

## x^3-x^2-x+3=0三根α,β,γ，求(α^2+α+1)(β^2+β+1)(γ^2+γ+1)

$$\displaystyle\left(\alpha^2+\alpha+1\right)\left(\beta^2+\beta+1\right)\left(\gamma^2+\gamma+1\right)$$

$$\displaystyle=\left(\alpha-\frac{1+\sqrt{3}i}{2}\right)\left(\alpha-\frac{1-\sqrt{3}i}{2}\right)\left(\beta-\frac{1+\sqrt{3}i}{2}\right)\left(\beta-\frac{1-\sqrt{3}i}{2}\right)\left(\gamma-\frac{1+\sqrt{3}i}{2}\right)\left(\gamma-\frac{1-\sqrt{3}i}{2}\right)$$

$$\displaystyle=\left[\left(\frac{1+\sqrt{3}i}{2}-\alpha\right)\left(\frac{1+\sqrt{3}i}{2}-\beta\right)\left(\frac{1+\sqrt{3}i}{2}-\gamma\right)\right]\cdot\left[\left(\frac{1-\sqrt{3}i}{2}-\alpha\right)\left(\frac{1-\sqrt{3}i}{2}-\beta\right)\left(\frac{1-\sqrt{3}i}{2}-\gamma\right)\right]$$

$$\displaystyle=f(\frac{1+\sqrt{3}i}{2})\cdot f(\frac{1-\sqrt{3}i}{2})=5\times5=25.$$

$$\displaystyle\Rightarrow \alpha ^2+ \alpha +1=\frac{5}{2-\alpha}$$

$$\displaystyle=\frac{5}{2-\alpha}\cdot\frac{5}{2-\beta}\cdot\frac{5}{2-\gamma}$$

$$\displaystyle=\frac{5^3}{f(2)}=25.$$

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