Question 121498

{{{343c^3+8}}}

since {{{343}}} is NOT a prime number, we can write it as:{{{ 343 = 7 * 7 * 7=7^3}}} and {{{8}}} as {{{2^3}}}

so, we will have:

{{{7^3c^3 + 2^3}}}

{{{(7c)^3 + 2^3}}}.........apply a sum of cubes rule:{{{(a +b)(a^2 - ab +b^2)}}}

{{{(7c + 2)( (7c)^2 - 14c + 4)}}}.........{{{(7c)^2 - 14c + 4)}}}.........
cannot be factored 

so, your answer is:
{{{highlight((7c + 2)(49c^2 - 14c + 4))}}}.........