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

To simplify the expression \( \frac{y-2}{y^{3}+1}-\frac{y-1}{y^{2}-y+1} \), let's first factor the denominators. The denominator \( y^3 + 1 \) can be factored using the sum of cubes formula: \[ y^3 + 1 = (y + 1)(y^2 - y + 1) \] Next, the second denominator \( y^2 - y + 1 \) does not factor further since it has no real roots. Now we can rewrite the expression with the common denominator. The common denominator will be \( (y + 1)(y^2 - y + 1) \). Rewriting the first term: \[ \frac{y-2}{y^3 + 1} = \frac{y-2}{(y + 1)(y^2 - y + 1)} \] Rewriting the second term: \[ \frac{y-1}{y^2 - y + 1} = \frac{(y-1)(y + 1)}{(y + 1)(y^2 - y + 1)} \] Now, we can combine the fractions: \[ \frac{y-2 - (y - 1)(y + 1)}{(y + 1)(y^2 - y + 1)} \] Expanding the numerator: \[ (y - 1)(y + 1) = y^2 - 1 \] So now we substitute: \[ \frac{y-2 - (y^2 - 1)}{(y + 1)(y^2 - y + 1)} \] This simplifies to: \[ \frac{y - 2 - y^2 + 1}{(y + 1)(y^2 - y + 1)} = \frac{-y^2 + y - 1}{(y + 1)(y^2 - y + 1)} \] So far we have: \[ \frac{-(y^2 - y + 1)}{(y + 1)(y^2 - y + 1)} \] The \(y^2 - y + 1\) in the numerator and denominator can be canceled out: \[ \frac{-1}{y + 1} \] Thus, the final simplified expression is: \[ \frac{-1}{y + 1} \]