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weiye 發表於 2012-3-3 11:04

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)

題目:多項式 \(x^3-x^2-x+3=0\) 的三根為 \(\alpha,\beta,\gamma\),

求 \(\displaystyle\left(\alpha^2+\alpha+1\right)\left(\beta^2+\beta+1\right)\left(\gamma^2+\gamma+1\right)\) 之值?



解答:

令 \(f(x)=x^3-x^2-x+3\),

經多項式的除法運算,

可得 \(f(x)=\left(x^2+x+1\right)\left(x-2\right)+5\)

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

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

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

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



另解:

令 \(f(x)=x^3-x^2-x+3\),

經多項式的除法運算,

可得 \(f(x)=\left(x^2+x+1\right)\left(x-2\right)+5\)

因此 \(f(\alpha)=0\Rightarrow\left(\alpha ^2+\alpha +1\right)\left(\alpha -2\right)+5=0\)

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


同理可得 \(\displaystyle \beta ^2+ \beta +1=\frac{5}{2-\beta}, \gamma ^2+ \gamma +1=\frac{5}{2-\gamma}\),

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

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

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

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