Factoring cubes/Solve by factoring:
1. w^3-2y^3

2. a^8-a^2b^6

3. 12qw^3-12q^4

Question

Factoring cubes/Solve by factoring:  
1. w^3-2y^3

2. a^8-a^2b^6

3. 12qw^3-12q^4

anonymous · Answer

$$w ^{3}-2y ^{3}  $$   $$a ^{8}-a ^{2}b ^{6}$$     $$12qw ^{3}-12q ^{4}$$

zepdrix · Answer

$$\large (a^3-b^3)=(a-b)(a^2+ab+b^2)$$
So this is the formula for difference of cubes, we'll need to use this to get through these :D

anonymous · Answer

Yes.  I have the formula.  Sounds good.  I was able to do some of the problems but I'm having a hard time with these three

zepdrix · Answer

Hmm the first one is a little annoying. Since we have that 2 there. It will make things a little ugly. Let's see if we can identify a and b.

anonymous · Answer

YES - I do not like the 2

zepdrix · Answer

$$\large a=w$$$$\large 2y^3=(\sqrt[3]{2}y)^3 =b^3$$$$\large b=\sqrt[3]{2} y$$
Understand what we did there? :o it was a little tricky.

zepdrix · Answer

We needed to rewrite the 2 as a cube. So we could think of it as the difference of cubes.

anonymous · Answer

It is a bit confusing but OK

zepdrix · Answer

sec, i'm trying to think if there is a better way we can write it :)

anonymous · Answer

What is confusing me is thinking of how to continue the formula with a radical cube

zepdrix · Answer

We continue the way we normally would, it just won't look very pretty :D|dw:1351391472114:dw|
$$\large =(w-\sqrt[3]{2}y)(w^2+\sqrt[3]{2}wy+y^2)$$