Answer
$-5x^{4a}-x^{2a}-1$
Work Step by Step
The expressions $
x^{6a}
$ and $
(x^{2a}+1)^3
$ are both perfect cubes (the cube root is exact). Hence, $
x^{6a}-(x^{2a}+1)^3
$ 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})^3-(x^{2a}+1)^3
\\\\=
[x^{2a}-(x^{2a}+1)][(x^{2a})^2+x^{2a}(x^{2a}+1)+(x^{2a}+1)^2]
\\\\=
[x^{2a}-x^{2a}-1][x^{4a}+x^{4a}+x^{2a}+(x^{4a}+2(x^{4a})(1)+(1)^2]
\\\\=
[x^{2a}-x^{2a}-1](x^{4a}+x^{4a}+x^{2a}+x^{4a}+2x^{4a}+1)
\\\\=
-1(5x^{4a}+x^{2a}+1)
\\\\=
-5x^{4a}-x^{2a}-1
.\end{array}