How do you factor completely $x^3 - 9x$ ?

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Answer 1 · 2016-08-01T04:49:12

$x^3-9x=x(x+3)(x-3)$

$x^3-9x$

= $x(x^2-9)$

= $x(x^2-3^2)$

Now using identity $a^2-b^2=(a+b)(a-b)$ , we get

$x(x^2-3^2)$

= $x(x+3)(x-3)$

How do you factor completely x^3 - 9x x^3 - 9x ?