Anyone here who can help me to solve these problems?

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1. (a^3 - a^2 + a - 1) / (a^3 + a^2 + a + 1)

Your first step would be to factor each of them. To factor these cubics, you need to use a technique called "grouping".

Factor the first two terms, and the last two terms, for each of the top and bottom.

[a^2(a - 1) + (a - 1)] / [a^2(a + 1) + (a + 1)]

Now, FACTOR each of them; notice the top has a common term of (a - 1), and the bottom has a common term of (a + 1).

[(a - 1) (a^2 + 1)] / [(a + 1) (a^2 + 1)]

Now, look what cancels on the top and the bottom; the common factor of (a^2 + 1). That leaves us with

(a - 1) / (a + 1)

2. (27y^3 - 1) / (27y^3 - 27y^2 + 9y - 1)

We can factor the numerator as a difference of cubes. A difference of cubes factors in the following manner:
(a^3 - b^3) = (a - b) (a^2 + ab + b^2). That means
(27y^3 - 1) = (3y - 1) (9y^2 + 3y + 1). Therefore, we get

[(3y - 1) (9y^2 + 3y + 1)] / (27y^3 - 27y^2 + 9y - 1)

Now, what we would *really* like is for (3y - 1) to be a factor of
(27y^3 - 27y^2 + 9y - 1), because that would lead to cancellation and simplfication. Let's actually do synthetic long division and determine if we get a remainder of 0. That is

(3y - 1) INTO (27y^3 - 27y^2 + 9y - 1)

I'm not going to show you the details, but I can tell you right now that we DO in fact get a remainder of 0, and a quotient of
9y^2 - 6y + 1. That means it factors into (3y - 1) (9y^2 - 6y + 1),
and our fraction is

[(3y - 1)(9y^2 + 3y + 1)] / [(3y - 1) (9y^2 - 6y + 1)]

Cancel the common factor,

(9y^2 + 3y + 1)/(9y^2 - 6y + 1)...Show more

1) (a^2+1)(a-1)/(a^2+1)(a+1) = (a-1)/(a+1)
2)I suppose the denominator is an equation
If you put 3y = z

z^3-3z^2-3z+1=0 which factored is (z+1)*(z^2-4z+1)

The numerator is (z+1)* (z^2+z+1)

so you expression is (z^2-4z+1)/(z2+z+1)

If it is different in any way it should help...Show more

1. Numerator=a^3-a^2+a-1
=a^2(a-1)+1(a-1)
=(a-1)(a^+1)
Denominator=a^2(a+1)+1(a+1)
=(a+1)(a^2+1)
Therefore,Numerator/Denominator
=(a-1)(a^2+1)/(a+1)(a^2+1)
=(a-1)/(a+1) ans
2. Numerator=(3y)^3-(1)^3
=(3y-1)(9y^2+3y+1) [following a^3-b^3=(a-b) (a^2-ab+b^2]
Denominator=27y^3-1-27y^2+9y
=(3y)^3-(1)^3-9y(3y-1)
=(3y-1)(9y^2+3y+1)-9y(3y-1)
=(3y-1)(9y^2+3y+1-9y)
=(3y-1)(9y^2-6y+1)
Therefore,Numerator/Denominator
=(9y^2+3y+1)/9y^2-6y+1)...Show more