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求實數a值範圍，使$x^3-3x+a=0$有:
(1)3個相異實根。
(2)重根

>$令f(x)=x^3-3x+a$
>$\Rightarrow f'(x)=3x^2-3$
>$\Rightarrow f'(\pm1)=0$
>
>(1)$f(x)=0$有3相異實根
>$\Rightarrow f(1)f(-1)<0$
>$\Rightarrow (a+2)(a-2)<0$
>$\Rightarrow -2<a<2$
>
>(2)$f(x)=0$有重根
>$\Rightarrow f(1)f(-1)=(a+2)(a-2)=0$
>$\Rightarrow a=\pm2$


