Answer
$-3x^{4a}+3x^{2a}-1$
Work Step by Step
The expressions $
(x^{2a}-1)^3
$ and $
x^{6a}
$ are both perfect cubes (the cube root is exact). Hence, $
(x^{2a}-1)^3-x^{6a}
$ is a $\text{
difference
}$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(x^{2a}-1)^3-(x^{2a})^3
\\\\=
[(x^{2a}-1)-x^{2a}][(x^{2a}-1)^2+(x^{2a}-1)(x^{2a})+(x^{2a})^2]
\\\\=
(x^{2a}-1-x^{2a})[(x^{4a}-2(x^{2a})(1)+(1)^2)+x^{4a}-x^{2a}+x^{4a}]
\\\\=
-1(x^{4a}-2x^{2a}+1+x^{4a}-x^{2a}+x^{4a})
\\\\=
-1(3x^{4a}-3x^{2a}+1)
\\\\=
-3x^{4a}+3x^{2a}-1
.\end{array}
