<s>[INST] If $a^4+64=0$ and $a^2 \ne-4a-8$, what is the value of $a^2-4a$? 
If $a^4+64=0$ and $a^2 \ne-4a-8$, what is the value of $a^2-4a$?

I tried to add $-16a^2$ to both sides and doing some algebra but it didnt help much.
 [/INST] Here's how to find the factorization, it is very similar to completing the square, we have $$\begin{align*}a^4 + 64 &= a^4 + 16a^2 + 64 - 16a^2\\ &= (a^2 + 8)^2 - 16a^2 \\ &= (a^2 - 4a + 8)(a^2 + 4a + 8)\end{align*}$$
See also http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity
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