How do you factor $x^{3} - 729$ $x^3-729$ ?

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How do you factor $x^{3} - 729$ $x^3-729$ ?

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Answer 1 · 2015-12-10T10:46:29

$(x - 9) (x^{2} + 9 x + 81)$ $(x-9)(x^2+9x+81)$

Explanation:

You have to observe that $729 = 9^{3}$ $729=9^3$ , and so $x^{3} - 729$ $x^3-729$ is a difference of cubes.

In general, you have that

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

and $(a^{2} + a b + b^{2})$ $(a^2+ab+b^2)$ is no longer factorizable (it is also known as "false square", because it resembles ${(a + b)}^{2}$ $(a+b)^2$ , but the mixed product is halved).

In your case, $a = x$ $a=x$ and $b = 9$ $b=9$ , so we have

$x^{3} - 9^{3} = (x - 9) (x^{2} + 9 x + 81)$ $x^3-9^3 = (x-9)(x^2+9x+81)$

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How do you factor $x^{3} - 729$ $x^3-729$ ?

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