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How do you (((y-y)/(3*y-3))+(1/(y-1)))/(y+1)/3 in partial fractions?

An expression to simplify:

The solution

You have entered [src]
/ y - y      1  \
|------- + -----|
|3*y - 3   y - 1|
|---------------|
\     y + 1     /
-----------------
        3        
$$\frac{\frac{1}{y + 1} \left(\frac{- y + y}{3 y - 3} + \frac{1}{y - 1}\right)}{3}$$
(((y - y)/(3*y - 3) + 1/(y - 1))/(y + 1))/3
Fraction decomposition [src]
-1/(6*(1 + y)) + 1/(6*(-1 + y))
$$- \frac{1}{6 \left(y + 1\right)} + \frac{1}{6 \left(y - 1\right)}$$
      1           1     
- --------- + ----------
  6*(1 + y)   6*(-1 + y)
General simplification [src]
     1     
-----------
  /      2\
3*\-1 + y /
$$\frac{1}{3 \left(y^{2} - 1\right)}$$
1/(3*(-1 + y^2))
Common denominator [src]
    1    
---------
        2
-3 + 3*y 
$$\frac{1}{3 y^{2} - 3}$$
1/(-3 + 3*y^2)
Trigonometric part [src]
        1         
------------------
3*(1 + y)*(-1 + y)
$$\frac{1}{3 \left(y - 1\right) \left(y + 1\right)}$$
1/(3*(1 + y)*(-1 + y))
Powers [src]
        1         
------------------
3*(1 + y)*(-1 + y)
$$\frac{1}{3 \left(y - 1\right) \left(y + 1\right)}$$
1/(3*(1 + y)*(-1 + y))
Assemble expression [src]
        1         
------------------
3*(1 + y)*(-1 + y)
$$\frac{1}{3 \left(y - 1\right) \left(y + 1\right)}$$
1/(3*(1 + y)*(-1 + y))
Rational denominator [src]
        1         
------------------
3*(1 + y)*(-1 + y)
$$\frac{1}{3 \left(y - 1\right) \left(y + 1\right)}$$
1/(3*(1 + y)*(-1 + y))
Combinatorics [src]
        1         
------------------
3*(1 + y)*(-1 + y)
$$\frac{1}{3 \left(y - 1\right) \left(y + 1\right)}$$
1/(3*(1 + y)*(-1 + y))
Combining rational expressions [src]
        1         
------------------
3*(1 + y)*(-1 + y)
$$\frac{1}{3 \left(y - 1\right) \left(y + 1\right)}$$
1/(3*(1 + y)*(-1 + y))
Numerical answer [src]
0.333333333333333/((1.0 + y)*(-1.0 + y))
0.333333333333333/((1.0 + y)*(-1.0 + y))