For this case we must factor the following expression:
[tex]27a ^ 3b ^ 9 + 8c ^ 18[/tex]
So, we rewrite the terms like:
[tex]27a ^ 3b ^ 9 = (3ab ^ 3) ^ 3\\8c ^ {18} = (2c ^ 6) ^ 3[/tex]
So, we have:
[tex](3ab ^ 3) ^ 3 + (2c^6) ^ 3 =[/tex]
Being both perfect cube terms, we factorize by applying the cube sum formula:
[tex](a ^ 3 + b ^ 3) = (a + b) (a ^ 2-ab + b ^ 2)[/tex]
Where:
[tex]a = 3ab ^ 3\\b = 2c ^ 6\\(3ab ^ 3 + 2c ^ 6) ((3ab ^ 3) ^ 2- (3ab ^ 3) (2c ^ 6) + (2c ^ 6) ^ 2) =\\(3ab ^ 3 + 2c ^ 6) (9a^2b ^ 6-6ab ^ 3c ^ 6 + 4c ^ {12})[/tex]
Answer:
[tex](3ab ^ 3 + 2c ^ 6) (9a2b ^ 6-6ab ^ 3c ^ 6 + 4c ^ 12)[/tex]
