Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set - Page 338: 43

$(t^2+4y^2)(t^4-4t^2y^2+16y^4)$

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $ t^6+64y^6 ,$ is \begin{array}{l} (t^2+4y^2)[ (t^2)^2-(t^2)(4y^2)+(4y^2)^2] \\\\= (t^2+4y^2)(t^4-4t^2y^2+16y^4) .\end{array}