Question 87273
Perform the indicated operation:
x^3-1/x^2+1 divided by 9x^2+9x+9/x^2-x
:

{{{((x^3-1))/((x^2+1))}}}
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{{{((9x^2+9x+9))/((x^2-x))}}}
:
(x^3-1) can be factored as the difference of cubes; factor out common terms:
{{{((x-1)(x^2+x+1))/((x^2+1))}}}
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{{{(9(x^2+x+1))/(x(x-1))}}}
:
When you divide fractions, invert the dividing fraction and multiply:
{{{((x-1)(x^2+x+1))/((x^2+1))}}} * {{{(x(x-1))/(9(x^2+x+1))}}}; cancel (x^2+x+1)
:
{{{((x-1))/((x^2+1))}}} * {{{(x(x-1))/9)}}} = {{{(x(x-1)^2)/(9(x^2+1))}}}; Thats about all you can do with it.
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