Factor a^6 + b^6.

Step-by-step solution

1. 1

   Choose the split that works for a sum. The exponent $6$ can be seen as $2\times3$ or $3\times2$. Reading it as squares, $\left(a^{3}\right)^{2}+\left(b^{3}\right)^{2}$, is a sum of squares and does not factor. Reading it as cubes does work:

   $a^{6}+b^{6}=\left(a^{2}\right)^{3}+\left(b^{2}\right)^{3}$

   The lesson is that a sum factors through an odd exponent — here the hidden odd exponent is the $3$.

2. 2

   Apply the sum-of-cubes identity $u^{3}+v^{3}=(u+v)\left(u^{2}-uv+v^{2}\right)$ with $u=a^{2}$ and $v=b^{2}$:

   $a^{6}+b^{6}=\left(a^{2}+b^{2}\right)\left(\left(a^{2}\right)^{2}-a^{2}b^{2}+\left(b^{2}\right)^{2}\right)$

3. 3

   Simplify the powers. Since $\left(a^{2}\right)^{2}=a^{4}$ and $\left(b^{2}\right)^{2}=b^{4}$:

   $a^{6}+b^{6}=\left(a^{2}+b^{2}\right)\left(a^{4}-a^{2}b^{2}+b^{4}\right)$

   The middle term is $-a^{2}b^{2}$, the product $uv$ — not $-2a^{2}b^{2}$, which would belong to a squared binomial instead.

4. 4

   Verify by expansion and numerically. Expanding gives $a^{6}-a^{4}b^{2}+a^{2}b^{4}+a^{4}b^{2}-a^{2}b^{4}+b^{6}=a^{6}+b^{6}$ ✓ — the four middle terms cancel in pairs. The identity holds at all $289$ integer pairs with $-8\le a,b\le8$ ✓. Spot check $a=2,b=1$: $64+1=65$ and $(5)(16-4+1)=5\cdot13=65$ ✓.

5. 5

   Note what cannot be pushed further over the integers. $a^{2}+b^{2}$ is a sum of squares and is irreducible over the reals. The quartic $a^{4}-a^{2}b^{2}+b^{4}$ also has no real linear factors — it equals $\left(a^{2}+b^{2}\right)^{2}-3a^{2}b^{2}$, and although that is a difference of squares, splitting it introduces $\sqrt3$, so it is not an integer factorisation. Hence the two-factor answer is complete over $\mathbb{Z}$.