# Simplifying: difference of two cubes

• Apr 7th 2009, 09:15 PM
Tenskypoo
Simplifying: difference of two cubes
$\displaystyle \frac {x^2 - 27}{x - 3}$
• Apr 7th 2009, 09:16 PM
Jhevon
Quote:

Originally Posted by Tenskypoo
$\displaystyle \frac {x^2 - 27}{x - 3}$

based on your previous questions, it makes me feel this should be $\displaystyle \frac {x^3 - 27}{x - 3}$ ...is that right?
• Apr 7th 2009, 09:22 PM
Tenskypoo
yup. I've gotten to where it is

(x+27)(x^2 - (x times 27))+ 27^2
__________________________________
x-3

• Apr 7th 2009, 09:25 PM
Jhevon
Quote:

Originally Posted by Tenskypoo
yup. I've gotten to where it is

(x+27)(x^2 - (x times 27))+ 27^2
__________________________________
x-3

nope

ok, so if it is $\displaystyle x^3 - 27$ on top. you would write that as $\displaystyle x^3 - 3^3$ and apply the (difference of two cubes) formula: $\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
• Apr 7th 2009, 09:30 PM
Tenskypoo
ah, because 3 x 3 x 3= 27

So if it was, x^4 - 16 it would be x^4 - 2^4 just as an example to see if i understand.
• Apr 7th 2009, 09:34 PM
Jhevon
Quote:

Originally Posted by Tenskypoo
ah, because 3 x 3 x 3= 27

So if it was, x^4 - 16 it would be x^4 - 2^4 just as an example to see if i understand.

ah, yeah, you could write it like that. but the formula i gave does not apply to that expression. but yes, that's the idea
• Apr 7th 2009, 09:35 PM
Tenskypoo
Quote:

Originally Posted by Jhevon
nope

ok, so if it is $\displaystyle x^3 - 27$ on top. you would write that as $\displaystyle x^3 - 3^3$ and apply the (difference of two cubes) formula: $\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

so would it be

x^3 - 3^3 = (x - 3)(x^2 + (x times 3) + 3^2)
• Apr 7th 2009, 09:36 PM
Jhevon
Quote:

Originally Posted by Tenskypoo
so would it be

x^3 - 3^3 = (x - 3)(x^2 + (x times 3) + 3^2)

yes, that's the numerator
• Apr 7th 2009, 09:38 PM
Tenskypoo
i'd put that over x-3, but cancel out the x-3 since it is in both the numer. and denom. therefore leaving me with

$\displaystyle x^2 + 3x + 9$

i may have done the part wrong, or it isn't simplified enough
• Apr 7th 2009, 09:39 PM
Jhevon
Quote:

Originally Posted by Tenskypoo
i'd put that over x-3, but cancel out the x-3 since it is in both the numer. and denom. therefore leaving me with

$\displaystyle x^2 + 3x + 9$

i may have done the part wrong, or it isn't simplified enough

that is fine
• Apr 8th 2009, 05:10 AM
stapel
Quote:

Originally Posted by Tenskypoo
yup. I've gotten to where it is

(x+27)(x^2 - (x times 27))+ 27^2
__________________________________
x-3

Make sure you learn and memorize the formula for factoring a difference of cubes before the next test! (Wink)