Question

Problema 39. Cu explicatie completa, daca puteti.
Multumesc!

Answer 1

$a^3 = 1 \Rightarrow a^3-1 = 0 \Rightarrow (a-1)(a^2+a+1) = 0 \\ \\ \Rightarrow a^2+a+1 = 0 \Rightarrow a^2 = -a-1\\ \\ (1-a)(1-a^2)(1-a^4)(1-a^5)=(1-a)(1-a^2)(1-a^3\cdot a)(1-a^3\cdot a^2) = \\ \\ = (1-a)(1-a^2)(1-a)(1-a^2) =\\ \\ = (1-a)^2(1-a^2)^2 = \Big[(1-a)(1-a^2)\Big]^2= \\ \\ = \Big[(1-a)\big(1-(-a-1)\big)\Big]^2 = \Big[(1-a)(2+a)\Big]^2= \\ \\ = (2-a-a^2)^2=\Big[2-a-(-a-1)\Big]^2 = (2-a+a+1)^2 = 3^2 = 9$