first, we can rewrite x^6 as (x^2)^3, and we can also rewrite 343 as 7^3.
now, we can apply the Sum of Cubes formula.
here is the sum of cubes formula, for future reference (if needed): \(x^3+y^3 = (x+y)(x^2-xy+y^2)\)
now to apply it to the current problem.
our x in this case is x^2, and our y would be 7.
so the factored form:
\(x^6+343 = (x^2+7)(x^4-7x^2+7^2)\)
= \((x^2+7)(x^4-7x^2+49)\), which is option D.
hope this helped! please let me know if you are confused about anything i did :)