How do you factor $x^{3} + 27 = 0$ $x^3+27=0$ ?

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How do you factor $x^{3} + 27 = 0$ $x^3+27=0$ ?

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Answer 1 · 2017-02-18T17:17:32

$(x + 3) (x^{2} - 3 x + 9) = 0$ $(x+3)(x^2-3x+9)=0$

Explanation:

We have a standard polynomial on LHS of the form $a^{3} + b^{3}$ $a^3+b^3$ , whose one factor is $(a + b)$ $(a+b)$ . So one factor is $(x + 3)$ $(x+3)$

$x^{3} + 27 = 0$ $x^3+27=0$

$\Leftrightarrow x^{3} + 3 x^{2} - 3 x^{2} - 9 x + 9 x + 27 = 0$ $hArrx^3+3x^2-3x^2-9x+9x+27=0$

or $x^{2} (x + 3) - 3 x (x + 3) + 9 (x + 3) = 0$ $x^2(x+3)-3x(x+3)+9(x+3)=0$

or $(x + 3) (x^{2} - 3 x + 9) = 0$ $(x+3)(x^2-3x+9)=0$

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How do you factor $x^{3} + 27 = 0$ $x^3+27=0$ ?

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