Ok guys...this college algebra stuff is kicking my butt! Here is the problem. I'll show it step by step of how my book has it:
1) 27t^3 - 64 =0
2) (3x)^3 - (4) ^3
3) (3x-4) (9x + 12x + 16) = 0

After that, the solve for the 3x-4, and the 9x+12x+16 (the latter is a quadratic equation using the quadratic formula). I totally understand how to work both of those, however what I'm stuck on is how they get from step 2 to step 3... I have a feeling once someone says it, I'm going to be like "duh" Can someone help me get to the "duh" point??

Question

Ok guys...this college algebra stuff is kicking my butt!  Here is the problem. I'll show it step by step of how my book has it:
 1) 27t^3 - 64 =0 
2) (3x)^3 - (4) ^3
3) (3x-4) (9x + 12x + 16)  = 0

After that, the solve for the 3x-4, and the 9x+12x+16 (the latter is a quadratic equation using the quadratic formula). I totally understand how to work both of those, however what I'm stuck on is how they get from step 2 to step 3... I have a feeling once someone says it, I'm going to be like "duh"   Can someone help me get to the "duh" point??

anonymous · Answer

1st one is a difference of 2 cubes

anonymous · Answer

(3t-4)(9t^2+12t+16)

anonymous · Answer

You want to know how to use the quadratic formula?

anonymous · Answer

No, I know how to use the quadratic formula. That's the easy part for me. I guess what I don't understand is finind the difference between the two squares

anonymous · Answer

No wait, I know how to find the difference. What I'm struggling on is how they get from step 2 to step 3. What is the process?

anonymous · Answer

$$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

anonymous · Answer

What step exactly do you mean?

anonymous · Answer

How do they get this:  (3x)^3 - (4) ^3 3)
to this:   (3x-4) (9x + 12x + 16) =

anonymous · Answer

Formula used is a^3 - b^3 = (a-b)(a^2 + a*b + b^2)

anonymous · Answer

Yes that is the difference of 2 cubes

cristiann · Answer

You use the formula
a^3-b^3=(a-b)(a^2+ab+b^2)
which you may easily verify by multiplying in the right-hand term

anonymous · Answer

Ok, I understand it is the difference of two cubes, what I don't understand is how they do it. Is it distributing?

anonymous · Answer

It's just a special rule

anonymous · Answer

Ok, having the formula is helping now

anonymous · Answer

The sum of 2 cubes would be:
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$

radar · Answer

The rule is$$x ^{3}-a ^{3}=(x-a)(x ^{2}+ax+a ^{2})$$

anonymous · Answer

thank you!!! my book doesn't say anything about using the formula. That's exactly what I was stuck on.

radar · Answer

You have to memorize it or just remember that there is factors of the difference of two perfect cubes.

radar · Answer

There is also one for the sum of two perfect cubes.  In this case it is the difference you are working with.

anonymous · Answer

Yes, I threw that in above in case you need it in the future

radar · Answer

Which I believe it will be needed.

anonymous · Answer

thank you, I wrote down both of them. Thanks guys!!