Question

计算[x+(1)/x][x^2+(1)/x^2][x^4+(1)/x^4][x^8+(1)/x^8][x^16+(1)/

Answer 1

[x+(1)/x][x^2+(1)/x^2][x^4+(1)/x^4][x^8+(1)/x^8][x^16+(1)/x^16](x^2-1)

=[x-(1)/x][x+(1)/x][x^2+(1)/x^2][x^4+(1)/x^4][x^8+(1)/x^8][x^16+(1)/x^16](x^2-1)/[x-(1)/x]

=[x^2-(1)/x^2][x^2+(1)/x^2][x^4+(1)/x^4][x^8+(1)/x^8][x^16+(1)/x^16](x^2-1)/[(x^2-1)/x]

=[x^4-(1)/x^4][x^4+(1)/x^4][x^8+(1)/x^8][x^16+(1)/x^16](x^2-1)* [x/(x^2-1)]

=[x^8-(1)/x^8][x^8+(1)/x^8][x^16+(1)/x^16]*x

=[x^16-(1)/x^16][x^16+(1)/x^16]*x

=x[x^32-(1)/x^32]