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x^3+ax^2-(a^2+1)x=0x(x^2+ax-a^2-1)=0x=0またはx^2+ax-a^2-1=0方程式x^2+ax-a^2-1=0Δ=a²-4(-a^2-1)=a²+4a²+4=5a²+4≥4>0だから、x^2+ax-a^2-1=0は2つの等しくない実数根を持つ。
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