I'm trying to factor $$a^6-b^6.$$
I have two options here: factor $a^6$ and $b^6$ either as cubes or as squares. I tried to factor them as cubes but had a single incorrect sign in the outcome.
$$a^6-b^6=(a^2)^3-(b^2)^3 = (a^2-b^2)((a^2)^2+a^2b^2+(b^2)^2)$$ Now factor out the power 2:
$$a^6-b^6=(a+b)(a-b)(a^2+ab+b^2)(a^2+ab+b^2)$$
The author of the exercise solves the exercise as follows: $$(a^3)^2-(b^3)^2 = (a^3+b^3)(a^3-b^3)$$ $$a^6 - b^6 = (a + b)(a^2 - ab + b^2)(a - b)(a^2 + ab + b^2)$$ $$a^6 - b^6 = (a + b)(a - b)(a^2 - ab + b^2)(a^2 + ab + b^2)$$
Notice the difference in sign of ab, once negative and once positive.
I fail to see the error in the way I solved it first.
Hint: $(a^4+a^2b^2+b^4)=(a^4+b^4+2a^2b^2-a^2b^2)=(a^2+b^2-ab)(a^2+b^2+ab)$.