math algebra problem?

what is the

8b^3-1

divided by

2b-1

???

Update:

wid details not just ans plz

Comments

  • 8b^3-1=(2b)^3-1=(2b-1)((2b)^2+2b+1)

    so the answer is

    (2b)^2+2b+1 or 4b^2+2b+1

  • You can put the first number over the second like a fraction and cancel out what you can.

    8b^3-1

    ---------

    2b-1

    so first you can cancel out the 2 and you are left with 4 on top

    b^3-1 and b-1 are both factors and can not be reduced any more.

    You are left with 4b^3-1/b-1

  • Ok, let's see how this one works

    dividing (8b^3-1)/(2b-1)

    = 4b^2+2b+1

    The problem does not ask for roots, so this is where we are stopping.

    Good Luck!

  • 8b^3-1 / 2b-1

    = 4b^2

  • 8b^3-1/2b-1 = (2b-1) ( 4b^2+2b+1)/2b-1, so that the answer is 4b^2+2b+1

  • I cannot do your homework for you ;) But you can try to multiply

    (x^2 + xy + y^2) * (x - y)

    this leads you to the well-known expression

    (x^3 - y^3) / (x - y) = x^2 + xy + y^2

    which gives the answer to your problem

  • ......... 4b^2-2b+1

    ......... |---------------------------

    2b-1 | 8b^3 + 0b^2 + 0b - 1

    ............8b^3 - 4b^2

    ......... --------------------------

    ................... -4b^2 + 0b - 1

    ................... -4b^2 + 2b

    ................... --------------------

    ............................... 2b - 1

    ............................... 2b - 1

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