Question 312922
The difference of two cubes is factored as follows:
{{{A^3-B^3 = (A-B)(A^2+AB+B^2)}}}, so...
{{{n^6-343 = (n^2)^3-(7)^3}}} and...
{{{(n^2)^3-(7)^3 = (n^2-7)(n^4+7n^2+49)}}}