How do I simplify $p^8-Q^8$? How do I simplify the expression $ p^8-q^8$?
Also, and explanation of how you got/why that's the answer would be great
 A: Before saying anything else:  first, I'm assuming the "$Q$" in the title and the "$q$" in the question are the same, so without ambiguity the question is to simplify $p^8 - q^8$; second, I assume by "simplify" the OP means "factor", though in agreement with the comment of timvermeulen, I'm not convinced that factoring really makes things any more simple in this case.  It's arguable either way, depending on one's intended use of the possible outcomes.  But I think factoring is consistent with the traditional/conventional/mainstream use of the word "simplify" in this context, so factoring it is.  Furthermore, I'm going to assume that "factor" here means "factor over the real numbers $\Bbb R$", so that all polynomials admitted will be required to have real coefficients.
These things being said:  the essential formula we will need here is this:  $a^2 - b^2 = (a - b)(a + b)$; this is quite easy to see, if one hasn't already:  
$(a - b)(a + b) = a(a + b) - b(a + b) = a^2 + ab - ba - b^2 = a^2 - b^2. \tag{1}$
Next, note that for any positive integer $m$,
$p^{2m} = (p^m)^2; \tag{2}$
the same is true for $q$, obviously:
$q^{2m} = (q^m)^2, \tag{3}$
so that taking $m = 4$ we have
$p^8 - q^8 = (p^4)^2 - (q^4)^2 = (p^4 -q^4)(p^4 + q^4); \tag{4}$
weallll, hmmmfff!  $p^4 + q^4$ looks sort of tough; so lets work on $(p^4 -q^4)$:  now taking $m = 2$ we get
$p^4 - q^4 = (p^2)^2 - (q^2)^2 = (p^2 - q^2)(p^2 + q^2), \tag{5}$
again using (1).  And again, $p^2 + q^2$ looks sort of challenging, but $p^2 - q^2$ we have already done in (1):
$p^2 - q^2 = (p - q)(p + q). \tag{6}$
Now we stick (6) into (5):
$p^4 - q^4 = (p^2)^2 - (q^2)^2 = (p - q)(p + q)(p^2 + q^2), \tag{7}$
and (7) into (4), yielding
$p^8 - q^8 = (p^4)^2 - (q^4)^2 = (p -q)(p + q)(p^2 + q^2)(p^4 + q^4), \tag{8}$
which is far as I know how to go in the direction of real factorization at present.  If anyone can take this further (of which I am a skeptic), then I'd like to hear about it. There are however a couple of interesting points to be made.  For instance, how do we know $p^2 + q^2$ can't be broken down into even simpler (e.g. of lower degree) factors?  Well, suppose we try to express $p^2 + q^2$ in terms of expressions of degree one, like $ap + bq$, where $a, b \in \Bbb R$.  If we could do this, we would get
$p^2 + q^2 = (ap + bq)(cp + dq), \tag{9}$
and if we distribute and multiply everything in (9) out we find
$p^2 + q^2 = acp^2 + (ad + bc)pq + bdq^2, \tag{10}$
and if we equate coefficients of like terms on either side we find that
$ac = bd = 1 \tag{11}$
and
$ad + bc = 0. \tag{12}$
(11) implies
$c = \frac{1}{a}, \tag{13}$
and
$d = \frac{1}{b}, \tag{14}$
and if we stick these into (12):
$\frac{a}{b} + \frac{b}{a} = 0, \tag{15}$
and if we multiply (15) by $ab$:
$a^2 + b^2 = 0,  \tag{16}$
which since $a, b \in \Bbb R$ means
$a = b = 0!  \tag{17}$
Whoa!  That won't work:  then $ap + bq = 0$ and everything collapses!  So we can't factor $p^2 + q^2$ like that, in any event.  A more advanced study might require
trying out various other possibilities to see where they might lead, but I'll betcha the answer is :  not very far.  UNLESS, of course, one allows the coefficients to be in the set of complex numbers $\Bbb C$.  Then some things work, like
$p^2 + q^2 = (p + iq)(p - iq), \tag{18}$
$p^4 + q^4 = (p^2 + iq^2)(p^2 - iq^2), \tag{19}$
and there is more, but I'll leave that for my readers to discover.  As far as real factorization is concerned, I'm stopping at (8) and will type no more for now.
So, my friends, 
Sayonara, 
Hope this helps!  Cheerio, 
and, as always,
Fiat Lux!!!
A: It is the difference of two squares.  One of the factors is again the difference of two squares.  And yet again.
