Factor x^5 + y^5.

Step-by-step solution

1. 1

   Check that a sum can factor at all. Sums of powers factor only when the exponent is odd. The reason is the root: $x=-y$ makes $x^{5}+y^{5}=-y^{5}+y^{5}=0$, so by the factor theorem $(x+y)$ must divide it. For an even exponent that substitution gives $2y^{4}\neq0$, which is why $x^{4}+y^{4}$ has no such factor.

2. 2

   Apply the sum-of-odd-powers identity. For odd $n$,

   $x^{n}+y^{n}=(x+y)\left(x^{n-1}-x^{n-2}y+x^{n-3}y^{2}-\cdots+y^{n-1}\right)$

   The signs alternate, starting positive and — because $n-1$ is even, so the number of terms $n$ is odd — ending positive.

3. 3

   Write out the $n=5$ case. The second factor has $5$ terms, with total degree $4$ in each:

   $x^{5}+y^{5}=(x+y)\left(x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}+y^{4}\right)$

4. 4

   Verify by expanding. Multiplying out, the $x$ distributes to give $x^{5}-x^{4}y+x^{3}y^{2}-x^{2}y^{3}+xy^{4}$ and the $y$ gives $x^{4}y-x^{3}y^{2}+x^{2}y^{3}-xy^{4}+y^{5}$. Every middle term cancels in pairs, leaving $x^{5}+y^{5}$ ✓ — the alternating signs exist precisely to force that telescoping. The identity was confirmed at all $289$ integer pairs with $-8\le x,y\le8$ ✓.

5. 5

   Confirm the quartic factor is irreducible over the integers. Treated as a quadratic in $x^{2}$ it has no rational structure, and it takes only positive values for real $x,y$ not both zero (for example at $x=y=1$ it is $1-1+1-1+1=1$). So this two-factor form is complete. Spot check $x=2,y=1$: $32+1=33$ and $(3)(16-8+4-2+1)=3\cdot11=33$ ✓.