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Homework Help: How to factorize(y + 3)^3 + 8

  1. Mar 8, 2008 #1
    [tex](y + 3)^3 + 8[/tex]

    my attempt has led me to


    but i doubt this is correct
  2. jcsd
  3. Mar 8, 2008 #2


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    Hi alpha01!

    If the answer isn't obvious, then simplify the question as much as possible, then try to solve it, then "un-simplify" it back.

    Put z = y + 3.

    Then the question is: factorise [tex]z^3 + 8[/tex].

    Can you do that? :smile:

    (Hint: when is it zero?)
  4. Mar 8, 2008 #3
    expanding and simplify the (y+3)^3:

    (y^2 + 6y + 9)(y + 3)

    so is the answer:

    (y^2 + 6y + 9) (y + 3 + 8)?
  5. Mar 8, 2008 #4
    Please post the solution and i will apply it to other examples
  6. Mar 8, 2008 #5


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    No no no … that is expanding but it's not simplifying! :frown:

    "Expanding" means "longer", and "simplifying" means "shorter"! :smile:

    [tex]z^3 + 8[/tex] is shorter and simpler than [tex](y + 3)^3 + 8[/tex].

    So: can you factorise [tex]z^3 + 8[/tex]? :smile:
  7. Mar 8, 2008 #6

    so the answer to the original question is:


    Last edited: Mar 8, 2008
  8. Mar 8, 2008 #7


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    Hint: can you see a number for which [tex]z^3\,+\,8\,=\,0[/tex]?

    (In other words, a root of [tex]z^3\,+\,8\,=\,0[/tex])

    Put "z - " in front of it, and that gives you a factor of [tex]z^3\,+\,8\,=\,0[/tex]! :smile:

    (if you still can't get it, I'll give you the answer)
  9. Mar 8, 2008 #8
  10. Mar 8, 2008 #9
    Solve for z:

    You need to think of this as

    [tex]z^3+x^3 = 0[/tex]

    Where x^3 = 8
    Last edited: Mar 8, 2008
  11. Mar 8, 2008 #10
    z = -6
  12. Mar 8, 2008 #11
    How are you getting z = -6?



  13. Mar 8, 2008 #12
    To solve this one, you should know this formula: a^3 + b^3 = (a + b)(a - ab + b). I hope that helps.
  14. Mar 8, 2008 #13

    thank you.. thats all i needed
  15. Mar 8, 2008 #14

    i meant to write -2

    i know how to solve this now anyway
  16. Mar 8, 2008 #15
    So, putting (y+3)^3 +8 in the form (a+b)(a^2 -ab+b^2):

    [tex]((y+3)+2) ((y+3)^2 -2(y+3)+4)[/tex]

    but i dont think this is completely factorized?
  17. Mar 9, 2008 #16

    Gib Z

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    I'm not trying to be rude, but from posts 1, 3 and 6 it seems to me you really need to go back and brush up on your algebra.
  18. Mar 9, 2008 #17


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    Well, you must then expand the two main brackets so that there's no brackets inside them!

    At that point, you've finished - it can't be factorised any further.

    [size=-2](Unless you know about complex numbers.)[/size]

    Though, just to check, you should then multiply it out again to make sure you get back to the original!

    Are you still worried about anything? :smile:
  19. Mar 9, 2008 #18
    I'm not trying to be rude, but posts 1, 3, 6 are all direct attempts of factorization. My general algebraic skills are fine (19/20 of a algebra quiz (no factorization questions) supports this).

    tiny-tim, thanks all solved.
    Last edited: Mar 9, 2008
  20. Mar 9, 2008 #19
    ok heres post #3

    (y+3)^3 expands to:

    (y^2 + 6y + 9)(y + 3)


    So can you please explain (with detailed solution) how this is apparently incorrect (and yes i know it can be expanded further i deliberately didn't expand it further.)

    and yes i know what "simplifying" is, i didnt simplify because i hadn't yet fully expanded it and that would take it out of the ()() form, which is obviously not going to factorize it.
    Last edited: Mar 9, 2008
  21. Mar 9, 2008 #20


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    Your original post was:
    It certainly isn't correct because you can't take the "8" inside one factor. That 8 would now be multiplied by the other "(y+ 3)" terms and it isn't in the original form.

    What you should have seen immediately was that (y+ 3)3+ 8= (y+3)3+ 23. Do you know a general formula for factoring a3+ b3? If you are asked to do a problem like this, I suspect it is introduced in this section!
  22. Mar 9, 2008 #21
    Simplify this and you will have your answer.
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