For this case we must factor the following expression:
[tex]x^{12}y{18}+1[/tex]
We rewrite [tex]x^{12}y^{18}[/tex]as [tex](x ^ 4y ^ 6) ^ 3:[/tex]
[tex](x ^ 4y ^ 6) ^ 3 + 1[/tex]
Being both perfect cube terms, it is factored by applying the cube sum formula:
[tex]a ^ 3 + b ^ 3 = (a + b) (a ^ 2-ab + b ^ 2)[/tex]
Where:
[tex]a = x ^ 4y ^ 6\\b = 1[/tex]
So:
[tex](x ^ 4y ^ 6 + 1) ((x ^ 4y ^ 6) ^ 2-x ^ 4y ^ 6 + 1 ^ 2) =\\(x ^ 4y ^ 6 + 1) (x ^ {8} y ^ {12} -x ^ 4y ^ 6 + 1)[/tex]
Answer:
Option B
