How do you factor $64 x^{3} + y^{3}$ $64x^3 + y^3$ ?

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How do you factor $64 x^{3} + y^{3}$ $64x^3 + y^3$ ?

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Answer 1 · 2018-05-03T15:22:58

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

Explanation:

This is an example of factoring perfect cubes.

$64 x^{3} + y^{3}$ $64x^3+y^3$

$64$ $64$ is a perfect cube $4 \cdot 4 \cdot 4$ $4*4*4$
$x^{3}$ $x^3$ is a perfect cube $x \cdot x \cdot x$ $x*x*x$
$y^{3}$ $y^3$ is a perfect cube $y \cdot y \cdot y$ $y*y*y$

The factor pattern for factoring perfect cubes is always

$(a + b) (a^{2} - a b + b^{2})$ $(a+b)(a^2-ab+b^2)$
or
$(a - b) (a^{2} + a b + b^{2})$ $(a-b)(a^2+ab+b^2)$

The signs follow the acronym SOAP

First sign is the SAME as the expression
Second sign is the OPPOSITE of the expression
Third sign is ALWAYS POSITIVE

For this problem
$a = 4 x$ $a =4x$ the root of $64 x^{3}$ $64x^3$
$b = y$ $b=y$ the root of $y^{3}$ $y^3$

So plugging in we get

$(4 x + y) ({(4 x)}^{2} - (4 x) (y) + y^{3})$ $(4x+y)((4x)^2 - (4x)(y) +y^3)$

$(4 x + y) (16^{2} - 4 x y + y^{3})$ $(4x+y)(16^2 - 4xy +y^3)$

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How do you factor $64 x^{3} + y^{3}$ $64x^3 + y^3$ ?

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