How do you factor $x^3-7$ ?

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Answer 1 · 2016-02-19T12:27:49

Factors are $(x−root(3)7)(x^2+root(3)7*x+(root(3)49))$

One well known identity is $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ .

Using this and as expressing $x^3−7$ as difference cubes, it is

$(x^3−(root(3)7)^3)$ - Hence factors are

$(x−root(3)7)(x^2+root(3)7*x+(root(3)7)^2)$ or

$(x−root(3)7)(x^2+root(3)7*x+(root(3)49))$

How do you factor x^3-7 x^3-7?