I understand that you need to factor this expression (you have "factor" tagged the question).

Noticing the third power of x and y I suggest to use the sum of cubes when factorize, such that:

`2x^3+16y^3 = (root(3) 2)^3*x^3 + (root(3) 16)^3*y^3`

Use the next formula:

`a^3 + b^3 = (a+b)(a^2 - ab + b^2)`

`(root(3) 2)^3*x^3 + (root(3) 16)^3*y^3 = (root(3) 2*x + root(3)16*y)(root(3) 4*x^2- root(3)(32)*(xy) + root(3) 256*y^2)`

**Hence, factorizing the sum yields `2x^3+16y^3 =(root(3) 2*x + root(3)16*y)(root(3) 4*x^2- root(3)(32)*(xy) + root(3) 256*y^2)` **

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