Factor completely: $x^{6} - 1$.

Factor completely: $x^{6} - 1$.
  1. $(x - 1)(x + 1)(x^{2} + x + 1)(x^{2} - x + 1)$
  2. $(x^{3} - 1)(x^{3} + 1)$
  3. $(x^{2} - 1)(x^{4} + x^{2} + 1)$
  4. $(x - 1)^{6}$

Solution

$x^{6}-1 = (x^{3}-1)(x^{3}+1) = (x-1)(x^{2}+x+1)(x+1)(x^{2}-x+1)$. The other options are correct factorisations but not complete; only option (a) is fully factored.

Asked in: IMO

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