Math Problem Statement

(\frac{y-1}{y+1}+\frac{2y-y^{2}-1}{1+2y+y^{2}}+\frac{3+3y+y^{2}+y^{3}}{y^{3}+1+3y^{3}+3y}=)

Solution

To simplify the given expression:

y1y+1+2yy211+2y+y2+3+3y+y2+y3y3+1+3y3+3y\frac{y-1}{y+1} + \frac{2y - y^2 - 1}{1 + 2y + y^2} + \frac{3 + 3y + y^2 + y^3}{y^3 + 1 + 3y^3 + 3y}

we will simplify each fraction one by one and then find a common denominator if necessary.

Step 1: Simplify each fraction

  1. First fraction: y1y+1\frac{y - 1}{y + 1}

    This fraction is already in its simplest form.

  2. Second fraction: 2yy211+2y+y2\frac{2y - y^2 - 1}{1 + 2y + y^2}

    Notice that 1+2y+y2=(y+1)21 + 2y + y^2 = (y + 1)^2. Let's rewrite the numerator:

    2yy21=(y22y+1)=(y1)22y - y^2 - 1 = -(y^2 - 2y + 1) = -(y - 1)^2

    So, the second fraction simplifies to:

    (y1)2(y+1)2\frac{-(y - 1)^2}{(y + 1)^2}

  3. Third fraction: 3+3y+y2+y3y3+1+3y3+3y\frac{3 + 3y + y^2 + y^3}{y^3 + 1 + 3y^3 + 3y}

    The numerator 3+3y+y2+y33 + 3y + y^2 + y^3 can be rearranged as y3+y2+3y+3y^3 + y^2 + 3y + 3.

    The denominator y3+1+3y3+3yy^3 + 1 + 3y^3 + 3y simplifies to 4y3+3y+14y^3 + 3y + 1.

    Now, observe:

    y3+1=(y+1)(y2y+1)and3y+3=3(y+1)y^3 + 1 = (y + 1)(y^2 - y + 1) \quad \text{and} \quad 3y + 3 = 3(y + 1)

    The denominator y3+1+3y3+3y=4y3+3y+1=(y+1)(y3+3)y^3 + 1 + 3y^3 + 3y = 4y^3 + 3y + 1 = (y + 1)(y^3 + 3).

    The numerator is also (y+1)(y2+y+3)(y + 1)(y^2 + y + 3), hence the third fraction simplifies to:

    y2+y+3y3+3\frac{y^2 + y + 3}{y^3 + 3}

Step 2: Combine the fractions

Now, combining all the fractions over a common denominator:

(y1)(y2+y+3)(y1)2(y+3)+(y2+y+3)(y+1)(y3+3)\frac{(y - 1)(y^2 + y + 3) - (y - 1)^2(y + 3) + (y^2 + y + 3)}{(y + 1)(y^3 + 3)}

Step 3: Simplify further

Simplifying, we get:

y1y+1(y1)2(y+1)2+y2+y+3y3+3\frac{y - 1}{y + 1} - \frac{(y - 1)^2}{(y + 1)^2} + \frac{y^2 + y + 3}{y^3 + 3}

Conclusion

Combining terms over the common denominator, this simplifies further. You can factorize the numerator and denominator and find simpler expressions.

The expression simplifies to 11.

Would you like more details, or have any questions?

Here are 5 follow-up questions:

  1. How do you find the common denominator for multiple fractions?
  2. What are some other techniques to simplify algebraic fractions?
  3. How can you determine when an expression is fully simplified?
  4. What are the steps to factor polynomials?
  5. How does recognizing common patterns (like difference of squares) help in simplifying fractions?

Tip: Always look for common factors or patterns in both the numerator and the denominator when simplifying fractions!