Question

(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)-1/8*3^64

Answer 1

$(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)-\frac {1}{8}*3^{64}=$

$=\frac{(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)}{3^2-1}-\frac {1}{8}*3^{64}= \frac{(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)}{8}-\frac {3^{64}}{8}= \frac{(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)}{8}-\frac {3^{64}}{8}= \frac{(3^{16}-1)(3^{16}+1)(3^{32}+1)}{8}-\frac {3^{64}}{8}= \frac{(3^{32}-1)(3^{32}+1)}{8}-\frac {3^{64}}{8}= \frac{3^{64}-1}{8}-\frac {3^{64}}{8}= \frac{3^{64}-1-3^{64}}{8}=-\frac{1}{8}$