40x^6 - 135
This was a problem on my last in-class test. I have a question regarding some of the last steps in how I worked it out. This isn't just considered prime, right? Even though none of the numbers are perfect squares. x^6 is also (x^3)^2, I know, but nothing else worked out.
So I took the GCD and got 5(8x^6 - 27). (8x^6 - 27) ends up being the difference of two cubes, right? So now I have 5 * (2x^3)^3 - (3)^3. I changed that according to rules and wrote the answer as 5(2x^3 - 3)(4x^9+6x^3+9).
I really think I got something mixed up, though, regarding the powers of the x's. What did I do wrong? Something simple, I know, in the multiplication. (As usual, I mess up because of something simple.) I appreciate your help very much!
This was a problem on my last in-class test. I have a question regarding some of the last steps in how I worked it out. This isn't just considered prime, right? Even though none of the numbers are perfect squares. x^6 is also (x^3)^2, I know, but nothing else worked out.
So I took the GCD and got 5(8x^6 - 27). (8x^6 - 27) ends up being the difference of two cubes, right? So now I have 5 * (2x^3)^3 - (3)^3. I changed that according to rules and wrote the answer as 5(2x^3 - 3)(4x^9+6x^3+9).
I really think I got something mixed up, though, regarding the powers of the x's. What did I do wrong? Something simple, I know, in the multiplication. (As usual, I mess up because of something simple.) I appreciate your help very much!