How do you factor $8 a^{3} - 27$ $8a^3-27$ ?

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Answer 1 · 2016-02-02T00:15:01

$(2 x - 3) (2 x^{2} + 6 x + 9)$ $(2x-3)(2x^2+6x+9)$

1) Decide factoring method

In the equation both $8$ $8$ and $27$ $27$ are cubes so we can use the Difference of Cubes method of factoring

2) Solve for variables

The formula for the Difference of Cubes method is:
$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3=(a-b)(a^2+ab+b^2)$

First we find $a$ $a$ :
$a^{3} = 8 a^{3}$ $a^3=8a^3$
$\sqrt[3]{a^{3}} = \sqrt[3]{8 a^{3}}$ $root(3)(a^3)=root(3)(8a^3)$
$a = 2 a$ $a=2a$

Then we find $b$ $b$ :
$b^{3} = 27$ $b^3=27$
$\sqrt[3]{b^{3}} = \sqrt[3]{27}$ $root(3)(b^3)=root(3)(27)$
$b = 3$ $b=3$

3) Fill in formula

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

${(2 x)}^{3} - 3^{3} = (2 x - 3) (2 x^{2} + (2 x \cdot 3) + 3^{2})$ $(2x)^3-3^3=(2x-3)(2x^2+(2x*3)+3^2)$

4) Simplify

$(2 x - 3) (2 x^{2} + 6 x + 9)$ $(2x-3)(2x^2+6x+9)$

How do you factor 8a^3-278a3−278a^3-27?