Respuesta :

5naka
let a=2018, we can rewrite the question into

[tex](a-1)(a^9+a^8+...+a^2+a+2)\\=(a-1)(a^9+a^8+...+a^2+a+1)+(a-1)\\=a^{10}-1+a-1[/tex]

substitute back a = 2018, obtained

[tex] {2018}^{10} + 2016[/tex]
well, there is something wrong here
UncleCeyth

Answer:

2016

Step-by-step explanation:

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