x^3-3x^2+3x-9 / x^4-81
i cant seem to get this answer right
assuming your function is ??
You need to factor both the numerator and denominator to determine any cancelling expressions.
Then the denominator will factor as follows
This can be further factored using complex numbers. Is this a requirement?
Do you know the factor theorem? You will need it to factor the numerator. Try to find where a is an integer factor of 9.
This is just using the difference of squares method on the denominator, which is that
This is just using that . You see how both can be divided by and likewise can be divided by 3? Apply that rule again on , where is the a and and 3 are the x and y respectively. Then, I just divided both sides of the fraction by .
The numerato you can factor by grouping. If you group x^3 and -3x^2 together and 3x and -9 together. You take the common factor from each and get
x^2(x-3) + 3(x-3)
then by the laws of grouping you can put them together and get (x^2 + 3)(x-3) and in the denominator you have (x-3)(x+3)(x^2 + 9) and then you can cross out the (x-3) on the top and bottom and you're left with:
(x^2 + 3)
(x+3)(x^2 + 9)