1. ## factor

3v^3-192

factor completely

2. Originally Posted by taemonique
3v^3-192

factor completely
Factor out a 3 from each term, then use the difference of cubes formula

$\displaystyle 3(v^3-64)$

remember that

$\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$

3. so what is the answer? because that isnt factored completely

4. Originally Posted by taemonique
so what is the answer? because that isnt factored completely
use the formula he gave you, you want to learn how to do it not just get the answer.

5. i did that much already. I dont see how to completly fctor it out

6. $\displaystyle 3V^3-192=3(V^3-64)=3(V^3-4^3)=?$
use this :
and finish this by yourself.

7. Forgive me if I am incorrect. Not completely sure with factorising but I chanced upon this.

Without using the $\displaystyle 4^3$ but using the technique showed I got

$\displaystyle 3v^3-192=...$then$\displaystyle (3v-64)(3v^2+(3v*64)+64^2)$

8. Originally Posted by Mukilab
Forgive me if I am incorrect. Not completely sure with factorising but I chanced upon this.

Without using the $\displaystyle 4^3$ but using the technique showed I got

$\displaystyle 3v^3-192=...$then$\displaystyle (3v-64)(3v^2+(3v*64)+64^2)$
this is incorrect. you need think against.

9. Probably because I didn't account for the $\displaystyle 4^3$

Can you account for my mistake?

I'll try again

then

$\displaystyle 3v^3-192=....$then$\displaystyle (3v-4^3)(3v^2+(3v4^3)+4^6)$

10. Originally Posted by Mukilab
Probably because I didn't account for the $\displaystyle 4^3$

Can you account for my mistake?

I'll try again

then

$\displaystyle 3v^3-192=....$then$\displaystyle (3v-4^3)(3v^2+(3v4^3)+4^6)$

You have been told exactly what to do. Go back. Read post #2. For your question, what is a? What is b? Substitute. Get the answer.

11. Finally?

$\displaystyle \Rightarrow 3v^3-192$

$\displaystyle 3v^9-64^9 \Rightarrow = (3v^3-64^3)(3v^6+3v^3(-64^3)-64^6)$

12. Originally Posted by Mukilab
Finally?

$\displaystyle \Rightarrow 3v^3-192$

$\displaystyle 3v^9-64^9 \Rightarrow = (3v^3-64^3)(3v^6+3v^3(-64^3)-64^6)$
How can this possibly be correct!? Do what I said in post #8. Show all your working - every single step, starting with what is a and what is b.

13. 3v^3-192

^^^
original

a=$\displaystyle 3v^3$
b=$\displaystyle -192$

$\displaystyle a^3=3v^3x^3=^9$, final answer is $\displaystyle 3v^9 = a^3$
$\displaystyle b^3=-192^3$

$\displaystyle 3v^9-192^3=(3v^3-192)(3v^6+(3v^3x(-192))-192^2)$

14. Sorry for the x's, I do not know how to input multiplication

15. Originally Posted by Mukilab
Finally?

$\displaystyle \Rightarrow 3v^3-192$

$\displaystyle 3v^9-64^9 \Rightarrow = (3v^3-64^3)(3v^6+3v^3(-64^3)-64^6)$
Originally Posted by Mukilab
Sorry for the x's, I do not know how to input multiplication
Originally Posted by Mukilab
3v^3-192

^^^
original

a=$\displaystyle 3v^3$
b=$\displaystyle -192$

$\displaystyle a^3=3v^3x^3=^9$, final answer is $\displaystyle 3v^9 = a^3$
$\displaystyle b^3=-192^3$

$\displaystyle 3v^9-192^3=(3v^3-192)(3v^6+(3v^3x(-192))-192^2)$
firstt this you the the number 192 you can you use calculator to see $\displaystyle 192=3*4^3$
and change this into the $\displaystyle 3v^3-192$so you have $\displaystyle 3v^3- 3*4^3$$\displaystyle \Longrightarrow 3(v^3-4^3)$
so $\displaystyle a=v, b=4$