I cannot find the last factor of this expression? I'm supposed to factor $x^8-y^8$ (the exponents are 8 for both if it is too difficult to see) as completely as possible. It is easy to factor this to $(x+y)(x-y)(x^2+y^2)(x^4+y^4)$. However, the book says "(Hint: there are 5 factors. Note that we sat real coefficients, not just integers)". What is this elusive 5th factor? Thanks!

Hint $\rm\,\ x^4 + y^4 = (x^2+y^2)^2 - (\sqrt{2}\, xy)^2 =\, \cdots\ $ (factor the difference of squares)
Remark $\ $ One can perform analogous factorizations in more exotic cases, and these prove very useful for factor factoring integers having such forms. $ $ For example, $ $ Aurifeuille, Le Lasseur and Lucas $ $ discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x)\, =\, C_n(x)^2\! - n\, x\,  D_n(x)^2\;$. These play a role in factoring integers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations.
$$\begin{eqnarray}
\rm x^4 + 2^2 &=\,&\rm  (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\
\rm \frac{x^6 + 3^2}{x^2 + 3} &=\,&\rm  (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\
\rm \frac{x^{10} - 5^5}{x^2 - 5} &=\,&\rm  (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\
\rm \frac{x^{12} + 6^6}{x^4 + 36} &=\,&\rm  (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\
\end{eqnarray}$$
For more on this and related topics see Sam Wagstaff's  introduction to the Cunningham Project. 
