How do you write $x^{3} - 1$ $x^3-1$ in factored form?

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Answer 1 · 2018-04-16T20:25:39

$x^{3} - 1 = (x - 1) (x^{2} + x + 1)$ $x^3 -1 = (x-1)(x^2 +x +1)$

This is a type of factorising called the the sum or difference of two cubes:

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3 - b^3 = (a-b)(a^2+ab +b^2)$

The sum of cubes is factored as:

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3 + b^3 = (a+b)(a^2-ab +b^2)$

In this case we have: $x^{3} - 1$ $x^3 -1$ so follow the rule above.

$x^{3} - 1 = (x - 1) (x^{2} + x + 1)$ $x^3 -1 = (x-1)(x^2 +x +1)$

How do you write x^3-1x3−1x^3-1 in factored form?