How do you factor $x^{3} - 8 y^{3}$ $x^3-8y^3$ ?

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How do you factor $x^{3} - 8 y^{3}$ $x^3-8y^3$ ?

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Answer 1 · 2015-04-09T12:56:55

We can write this expression as $x^{3} - {(2 y)}^{3}$ $x^3-(2y)^3$

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

In $x^{3} - {(2 y)}^{3}$ $x^3-(2y)^3$ ,
$a = x$ $a=x$
$b = 2 y$ $b=2y$

$x^{3} - {(2 y)}^{3}$ $x^3-(2y)^3$
$= (x - 2 y) \cdot (x^{2} + (x \cdot 2 y) + {(2 y)}^{2})$ $= (x-2y)*(x^2+(x*2y)+(2y)^2)$
$= (x - 2 y) \cdot (x^{2} + 2 x y + 4 y^{2})$ $= color(green)( (x-2y)*(x^2+2xy+4y^2)$

As we cannot factorize any of the factors further, we can say that we have factorised $x^{3} - 8 y^{3}$ $x^3-8y^3$ completely.

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How do you factor $x^{3} - 8 y^{3}$ $x^3-8y^3$ ?

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