How do you factor $(x^{3} - 64)$ $(x^3 - 64)$ ?

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How do you factor $(x^{3} - 64)$ $(x^3 - 64)$ ?

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Answer 1 · 2018-03-30T07:56:40

$(x - 4) (x^{2} + 4 x + 16)$ $(x-4)(x^2+4x+16)$

Explanation:

$x^{3} - 64 \leftarrow is a difference of cubes$ $x^3-64larrcolor(blue)"is a difference of cubes"$

$∙ x a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)$

$here a = x and b = 4 \to {(4)}^{3} = 64$ $"here "a=x" and "b=4to(4)^3=64$

$\Rightarrow x^{3} - 64 = (x - 4) (x^{2} + 4 x + 16)$ $rArrx^3-64=(x-4)(x^2+4x+16)$

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How do you factor $(x^{3} - 64)$ $(x^3 - 64)$ ?

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How do you factor $(x^{3} - 64)$ $(x^3 - 64)$ ?