# How do you factor 54x^3 - 2y^3?

Mar 30, 2018

$2 \left(3 x - y\right) \left(9 {x}^{2} + 3 x y + {y}^{2}\right)$

#### Explanation:

When factoring,
Step 1: Always factor out a GCF first, if possible.

In this particular problem, we can factor out 2:
$54 {x}^{3} - 2 {y}^{3}$ = $2 \left(27 {x}^{3} - {y}^{3}\right)$

Step 2: Count the number of terms:

—if 2 terms, check if problem is:

difference of squares $\left({a}^{2} - {b}^{2}\right)$, which factors into $\left(a - b\right) \left(a + b\right)$

difference of cubes $\left({a}^{3} - {b}^{3}\right)$, which factors into $\left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

sum of cubes $\left({a}^{3} + {b}^{3}\right)$, which factors into $\left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

(27x^3-y^3) is a difference of cubes:
$27 {x}^{3}$ = ${\left(3 x\right)}^{3}$
${y}^{3}$ = ${\left(y\right)}^{3}$

Plugging these values into the difference of cubes formula:
$\left({a}^{3} - {b}^{3}\right)$ = $\left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$
((3x)^3-((y)^3) = $\left(\left(3 x\right) - \left(y\right)\right) \left({\left(3 x\right)}^{2} + \left(3 x\right) \left(y\right) + {\left(y\right)}^{2}\right)$
$27 {x}^{3} - {y}^{3}$ = $\left(3 x - y\right) \left(9 {x}^{2} + 3 x y + {y}^{2}\right)$

Putting it all together:

$54 {x}^{3} - 2 {y}^{3}$ = $2 \left(27 {x}^{3} - {y}^{3}\right)$
$54 {x}^{3} - 2 {y}^{3}$ = $2 \left(3 x - y\right) \left(9 {x}^{2} + 3 x y + {y}^{2}\right)$