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quickstudent

  • one year ago

Can someone please check this for me? Factoring Sum of Cubes

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  1. quickstudent
    • one year ago
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  2. hartnn
    • one year ago
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    3a is incorrect

  3. hartnn
    • one year ago
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    which makes 3b and 3c also incorrect

  4. quickstudent
    • one year ago
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    Ok, let me review it a moment...

  5. hartnn
    • one year ago
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    sure, i was about to say, try to find the error on your own

  6. hartnn
    • one year ago
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    and the same mistake is done in 4. you correct one, you correct all!! :)

  7. quickstudent
    • one year ago
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    Hmmm, I don't see what's wrong there. Can you explain?

  8. hartnn
    • one year ago
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    length = 2t its cube = \( (2t)^3 = 8t^3 \)

  9. hartnn
    • one year ago
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    basically you forgot the brackets :P

  10. quickstudent
    • one year ago
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    Oh, okay. I'll try doing those again then :)

  11. hartnn
    • one year ago
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    sure, all 3 a,b,c and 4 a,b,c needs that correction. let me know when you're done, i'll review it again.

  12. quickstudent
    • one year ago
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    Here it is now, corrected :) @hartnn

  13. hartnn
    • one year ago
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    lets come to 3b now. \((2t)^3 + 12^3 = (2t+12)((2t)^2 - (2t)\times 12) +12^2 \) makes sense?

  14. hartnn
    • one year ago
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    i missed the last bracket \((2t)^3 + 12^3 = (2t+12)((2t)^2 - (2t)\times 12 +12^2)\)

  15. hartnn
    • one year ago
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    got that^^ ?

  16. quickstudent
    • one year ago
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    OK, I get it.

  17. hartnn
    • one year ago
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    so your answer for 3b should be \((2t+12)(4t^2 -24t +144)\)

  18. hartnn
    • one year ago
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    now for 3c, its (2t)^3 not (8t)^3 ... so (2*7)^3 + 12^3 = 14^3 +12^3 = .........

  19. hartnn
    • one year ago
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    same thing goes for 4b and 4c :)

  20. hartnn
    • one year ago
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    \((2t+12)(4t^2 -24t +144)\) ^^ that can be further factorized \(8(t + 6)(t^2 - 6t + 36)\)

  21. quickstudent
    • one year ago
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    OK, now I've correct 3 and 4 again.

  22. hartnn
    • one year ago
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    typing error in 4b

  23. hartnn
    • one year ago
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    everything else is correct :)

  24. quickstudent
    • one year ago
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    I've corrected that typo:) Thanks for helping me:)

  25. hartnn
    • one year ago
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    welcome ^_^

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