Let s = (3+1/3) (3 2+1/3 2) (3 4+1/3 4) (3 8+1/3 8).

Reuse the square difference formula

(3- 1/3)s=(3- 1/3)(3+ 1/3)(3^2+ 1/3^2)(3^4+ 1/3^4)(3^8+ 1/3^8)(3^ 16+ 1/3^ 16)

=(3^2- 1/3^2)(3^2+ 1/3^2)(3^4+ 1/3^4)(3^8+ 1/3^8)(3^ 16+ 1/3^ 16)

=(3^4- 1/3^4)(3^4+ 1/3^4)(3^8+ 1/3^8)(3^ 16+ 1/3^ 16)

=(3^8- 1/3^8)(3^8+ 1/3^8)(3^ 16+ 1/3^ 16)

=(3^ 16- 1/3^ 16)(3^ 16+ 1/3^ 16)

=(3^32- 1/3^32)

Therefore, s = (3 32-1/3 32)/(3-1/3) = 3 (3 32-1/3 32)/8 = (3/8) (3 32-65438).