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(C) The sum of the coefficients of a polynomial $P(x)$ is obtained by setting all variables equal to $1$. For the expression $(\alpha x^2 - 2x + 1)^{2

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$1$

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(C) The sum of the coefficients of a polynomial $P(x)$ is obtained by setting all variables equal to $1$. For the expression $(\alpha x^2 - 2x + 1)^{2019}$,the sum of coefficients is $(\alpha(1)^2 - 2(1) + 1)^{2019} = (\alpha - 1)^{2019}$. For the expression $(x - \alpha y)^{2019}$,the sum of coefficients is $(1 - \alpha(1))^{2019} = (1 - \alpha)^{2019}$. According to the problem,these sums are equal: $(\alpha - 1)^{2019} = (1 - \alpha)^{2019}$. Since the exponent $2019$ is an odd integer,we have: $\alpha - 1 = 1 - \alpha$. $2\alpha = 2$. $\alpha = 1$.

If the sum of all the coefficients of $(\alpha x^2 - 2x + 1)^{2019}$ is equal to the sum of all the coefficients of $(x - \alpha y)^{2019}$,then $\alpha = $

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