Key idea. Use a difference of squares, followed by two cube identities.

1. First,

   \[x^6-1=(x^3-1)(x^3+1).\]

2. Factor each cubic:

   \[\begin{aligned}x^3-1&=(x-1)(x^2+x+1),\\x^3+1&=(x+1)(x^2-x+1).\end{aligned}\]

3. Combining the factors gives

   \[x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1).\]

4. Each quadratic has discriminant \(1-4=-3\), so neither has real roots. In particular, neither splits into linear factors with integer coefficients.

Remember. After one factorisation, inspect each remaining factor for another useful identity.