a) (x-3)3 = 3-x
<=> (x-3)3 + (x-3) = 0
<=> (x-3)[(x-3)2+1] = 0
<=> \(\left\{{}\begin{matrix}x-3=0\\\left(x-3\right)^2=-1\left(vl\right)\end{matrix}\right.\)
<=> x= 3
b) (x-2017)3 + 2017 = x
<=> (x-2017)3 + 2017 - x =0
<=> (x-2017)3 - (x-2017) = 0
<=> (x-2017)[(x-2017)2-1)] = 0
<=> \(\left\{{}\begin{matrix}x-2017=0\\\left(x-2017\right)^2=1\end{matrix}\right.\)
,=> \(\left\{{}\begin{matrix}x=2017\\x=2018\\x=2016\end{matrix}\right.\)
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