解：?$(2)$?原式?$=\frac 12×(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$?
?$=\frac 12×(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$?
?$=\frac 12×(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$?
?$=\frac 12×(3^8-1)(3^8+1)(3^{16}+1)$?
?$=\frac 12×(3^{16}-1)(3^{16}+1)$?
?$=\frac 12×(3^{32}-1)$?
?$=\frac {3^{32}-1}2$?
