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??

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- anonymous

- jamiebookeater

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- anonymous

1st one is a difference of 2 cubes

- anonymous

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

- anonymous

You want to know how to use the quadratic formula?

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## More answers

- anonymous

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

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

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

- anonymous

What step exactly do you mean?

- anonymous

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

- anonymous

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

- anonymous

Yes that is the difference of 2 cubes

- cristiann

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

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

- anonymous

It's just a special rule

- anonymous

Ok, having the formula is helping now

- anonymous

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

- radar

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

- anonymous

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

- radar

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

- radar

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

- anonymous

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

- radar

Which I believe it will be needed.

- anonymous

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

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