Question

What is the factored form of X^3-1

Answer 1

Keeping in mind that

1² = 1
1³ = 1
1⁴ = 1
1¹⁰⁰⁰⁰⁰⁰ = 1

$\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) \\\\ a^3-b^3 = (a-b)(a^2+ab+b^2)\\\\ -------------------------------\\\\ x^3-1\implies x^3-1^3\implies (x-1)(x^2+1x+1^2) \\\\\\ (x-1)(x^2+x+1)$

Answer 2

Notice how both terms are perfect cubes.  You can factor using the difference of cubes formula:

$a^3-b^3=(a-b)(a^2+ab+b^2) \\ \\ x^3-1 = x^3- 1^3 = (x-1)(x^2+x+1)$

The factored form of  x³ - 1 is (x - 1)(x² + x + 1).