x^3-1/x^2+1 divided by 9x^2+9x+9/x^2-x?

Question

I am still having problems with the division on my algebrator
can someone help?
I would like to see the steps.. it really helps

Answer 1

Factor each if possible; x^3-1 = (x-1)(x^2 + x + 1); 9x^2 + 9x + 9 = 9(x^2 + x + 1); x^2 - x = x(x-1)

Flip the second fraction and then see which factors cancel. The (x^2 + x + 1) cancels; nothing else does (unless you typed x^2 +1 when it should be x^2 -1)

So you get (x-1)(x)(x-1) on top and (x^2+1)(9) on bottom

Answer 2

((x^3 - 1)/(x^2 + 1))/((9x^2 + 9x + 9)/(x^2 - x))
((x^3 - 1)/(x^2 + 1)) * ((x^2 - x)/(9x^2 + 9x + 9))
((x^3 - 1)(x^2 - x))/((9x^2 + 9x + 9)(x^2 + 1))
(((x - 1)(x^2 + x + 1)(x(x - 1))) / (9(x^2 + x + 1)(x^2 + 1))

as you can see, there is a (x^2 + x + 1) on top and bottom, so unless you mistyped anything, those are the only ones you can take out.

(x(x - 1)(x - 1)) / (9(x^2 + 1))
(x(x^2 - x - x + 1))/(9x^2 + 1)
(x(x^2 - 2x + 1))/(9x^2 + 1)

ANS :

(x^3 + 2x^2 + x)/(9x^2 + 1)

Answer 3

x^3-1/x^2+1 divided by 9x^2+9x+9/x^2-x
is the same as x^3-1/x^2+1 multiplied by
x^2-x/9x^2+9x+9
so the numerator of the solution is x^3-1 times x^2-x
and the denominator is x^2+1 times 9x^2+9x+9

numerator: ( x^3-1)(x^2-x)=x^5-x^4-x^2-x

denominator: (x^2+1)(9x^2+9x+9)=
9x^4+9x^3+9x^2+9x^2+9x+9 =
9x^4+9x^3+18x^2+9x+9

so the answer is
x^5-x^4-x^2-x/9x^4+9x^3+18x^2+9x+9

Answer 4

hello baby . i really want to help but i can't send the steps by text . so if you want give me your email address and i will send you a photo contain the steps . and also in the last part of your question is it ...+9/x^(2-x) or ...+9/x^2 - x ??