$P(x)=x^4+4px^3-4qx-1$有重根的条件$(p+q)^{2/3}-(p-q)^{2/3}=1$

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观察到 $P(0)=-1\neq0$，所以任何根 $\alpha$ 都 $\neq0$$$P'(x)=4x^3+12px^2-4q$$解 $P'(\alpha)=0$，$$q=\alpha^3+3p\alpha^2$$代入 $P(\alpha)=0$$$\alpha^4+4p\alpha^3-4q\alpha-1=0\;\Longrightarrow\;3\alpha^4+8p\alpha^3+1=0\;\Longrightarrow\;
8p=-\frac{3\alpha^4+1}{\alpha^3}=-3\alpha-\frac1{\alpha^3}
$$
因此
$$8p=-3\alpha-\alpha^{-3}$$$$8q=-\alpha^3-3\alpha^{-1}$$$$
8(p+q)=-(\alpha^3+3\alpha+3\alpha^{-1}+\alpha^{-3})= -(\alpha+\tfrac1\alpha)^3$$$$8(p-q)=\alpha^3-3\alpha+3\alpha^{-1}-\alpha^{-3}= (\alpha-\tfrac1\alpha)^3.$$
消$\alpha$恰好是$(p+q)^{2/3} - (p-q)^{2/3}=1$，等价于 $P(x)$ 的判别式为零？