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anonymous

  • 4 years ago

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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  1. anonymous
    • 4 years ago
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    1st one is a difference of 2 cubes

  2. anonymous
    • 4 years ago
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    (3t-4)(9t^2+12t+16)

  3. anonymous
    • 4 years ago
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    You want to know how to use the quadratic formula?

  4. anonymous
    • 4 years ago
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    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

  5. anonymous
    • 4 years ago
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    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?

  6. anonymous
    • 4 years ago
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    \[\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

  7. anonymous
    • 4 years ago
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    What step exactly do you mean?

  8. anonymous
    • 4 years ago
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    How do they get this: (3x)^3 - (4) ^3 3) to this: (3x-4) (9x + 12x + 16) =

  9. anonymous
    • 4 years ago
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    Formula used is a^3 - b^3 = (a-b)(a^2 + a*b + b^2)

  10. anonymous
    • 4 years ago
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    Yes that is the difference of 2 cubes

  11. cristiann
    • 4 years ago
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    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

  12. anonymous
    • 4 years ago
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    Ok, I understand it is the difference of two cubes, what I don't understand is how they do it. Is it distributing?

  13. anonymous
    • 4 years ago
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    It's just a special rule

  14. anonymous
    • 4 years ago
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    Ok, having the formula is helping now

  15. anonymous
    • 4 years ago
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    The sum of 2 cubes would be: \[a^3+b^3=(a+b)(a^2-ab+b^2)\]

  16. radar
    • 4 years ago
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    The rule is\[x ^{3}-a ^{3}=(x-a)(x ^{2}+ax+a ^{2})\]

  17. anonymous
    • 4 years ago
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    thank you!!! my book doesn't say anything about using the formula. That's exactly what I was stuck on.

  18. radar
    • 4 years ago
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    You have to memorize it or just remember that there is factors of the difference of two perfect cubes.

  19. radar
    • 4 years ago
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    There is also one for the sum of two perfect cubes. In this case it is the difference you are working with.

  20. anonymous
    • 4 years ago
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    Yes, I threw that in above in case you need it in the future

  21. radar
    • 4 years ago
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    Which I believe it will be needed.

  22. anonymous
    • 4 years ago
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    thank you, I wrote down both of them. Thanks guys!!

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