Factor x^10 - y^10 completely.

Step-by-step solution

1. 1

   Choose the split that keeps every factor rational. Since $10=2\times5$, the exponent can be halved either way. Taking it as a difference of squares first is best, because $a^{2}-b^{2}$ splits into two factors immediately:

   $x^{10}-y^{10}=\left(x^{5}\right)^{2}-\left(y^{5}\right)^{2}=\left(x^{5}-y^{5}\right)\left(x^{5}+y^{5}\right)$

   (Starting instead from $\left(x^{2}\right)^{5}-\left(y^{2}\right)^{5}$ reaches the same end but needs more regrouping.)

2. 2

   Factor the difference of fifth powers. The general identity is $a^{n}-b^{n}=(a-b)\left(a^{n-1}+a^{n-2}b+\cdots+b^{n-1}\right)$, so with $n=5$ every term has a $+$ sign:

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

3. 3

   Factor the sum of fifth powers. A sum of odd powers also factors, with alternating signs in the long factor:

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

   This works only because $5$ is odd — $x^{4}+y^{4}$, by contrast, has no factorisation over the integers.

4. 4

   Assemble the complete factorisation.

   $x^{10}-y^{10}=(x-y)(x+y)\left(x^{4}+x^{3}y+x^{2}y^{2}+xy^{3}+y^{4}\right)\left(x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}+y^{4}\right)$

   The degrees add correctly: $1+1+4+4=10$ ✓. Both quartic factors are irreducible over the integers, so this is fully factored.

5. 5

   Verify numerically. Expanding the product at all $169$ integer pairs $(x,y)$ with $-6\le x,y\le6$ reproduces $x^{10}-y^{10}$ exactly ✓. A quick spot check with $x=2,y=1$: the left side is $1024-1=1023$, and the right side is $(1)(3)(16+8+4+2+1)(16-8+4-2+1)=1\cdot3\cdot31\cdot11=1023$ ✓.