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

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Answer 1 · 2016-02-19T12:27:49

Factors are $(x - \sqrt[3]{7}) (x^{2} + \sqrt[3]{7} \cdot x + (\sqrt[3]{49}))$ $(x−root(3)7)(x^2+root(3)7*x+(root(3)49))$

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

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

$(x^{3} - {(\sqrt[3]{7})}^{3})$ $(x^3−(root(3)7)^3)$ - Hence factors are

$(x - \sqrt[3]{7}) (x^{2} + \sqrt[3]{7} \cdot x + {(\sqrt[3]{7})}^{2})$ $(x−root(3)7)(x^2+root(3)7*x+(root(3)7)^2)$ or

$(x - \sqrt[3]{7}) (x^{2} + \sqrt[3]{7} \cdot x + (\sqrt[3]{49}))$ $(x−root(3)7)(x^2+root(3)7*x+(root(3)49))$

How do you factor x3−7 x^3-7?