A telescoping product

$(x+y)(x^2+y^2)(x^4+y^4)(x^8+y^8) = \dfrac{x^2-y^2}{x-y} (x^2+y^2)(x^4+y^4)(x^8+y^8)$ $= \dfrac{x^4-y^4}{x-y} (x^4+y^4)(x^8+y^8) = \dfrac{x^8-y^8}{x-y}(x^8+y^8)$ $= \dfrac{x^{16}-y^{16}}{x-y}$ $x=3$ and $y=2$

It is $\boxed{\dfrac{3^{16}-2^{16}}{3-2} = 3^{16}-2^{16}}$

To solve this problem, simply multiple the entire expression by 1. 1 is 3-2, so 1*(3+2)(3^2+2^2)(3^4+2^4)(3^8+2^8) is equal to (3-2)(3+2)(3^2+2^2)(3^4+2^4)(3^8+2^8), which is equal to (3^2-2^2)(3^2+2^2)(3^4+2^4)(3^8+2^8), since a^2-b^2 = (a-b)(a+b). Using this rule, we get that (3^2-2^2)(3^2+2^2) = (3^4-2^4), and (3^4-2^4)(3^4+2^4) = (3^8-2^8), and finally (3^8-2^8)(3^8+2^8) = 3^16-2^16.