How do you factor completely $64 + a^{3}$ $64 + a^3$ ?

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How do you factor completely $64 + a^{3}$ $64 + a^3$ ?

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Answer 1 · 2016-07-13T19:43:50

$a^{3} + 64 = (a + 4) (a^{2} - 4 a + 16)$ $a^3+64=(a+4)(a^2-4a+16)$

Explanation:

The expression $a^{3} + 64 = a^{3} + 4^{3}$ $a^3+64=a^3+4^3$ is the sum of two cubes (3rd powers). In general, if you have an arbitrary sum of two cubes, say $x^{3} + y^{3}$ $x^3+y^3$ , it can be factored as:

$x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$ $x^3+y^3=(x+y)(x^2-xy+y^2)$ . You should check this by expansion (multiplication) of the right-hand side.

Now apply this formula to the problem at hand by substituting $x = a$ $x=a$ and $y = 4$ $y=4$ .

The difference of two cubes can also be factored, and the formula for the sum of two cubes can be used to do it:

$x^{3} - y^{3} = x^{3} + {(- y)}^{3}$ $x^3-y^3=x^3+(-y)^3$

$= (x + (- y)) (x^{2} - x (- y) + {(- y)}^{2})$ $=(x+(-y))(x^2-x(-y)+(-y)^2)$

$= (x - y) (x^{2} + x y + y^{2})$ $=(x-y)(x^2+xy+y^2)$ .

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How do you factor completely $64 + a^{3}$ $64 + a^3$ ?

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