Find $ \left(1-y\right)^3 $ using formula

$$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$

where $ A = 1 $ and $ B = y $.

$$ \left(1-y\right)^3 = 1^3-3 \cdot 1^2 \cdot y + 3 \cdot 1 \cdot y^2-y^3 = 1-3y+3y^2-y^3 $$

Find $ \left(1-y\right)^2 $ using formula.

$$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ 1 } $ and $ B = \color{red}{ y }$.

$$ \begin{aligned}\left(1-y\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot y + \color{red}{y^2} = 1-2y+y^2\end{aligned} $$

Find $ \left(1-y\right)^3 $ using formula

$$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$

where $ A = 1 $ and $ B = y $.

$$ \left(1-y\right)^3 = 1^3-3 \cdot 1^2 \cdot y + 3 \cdot 1 \cdot y^2-y^3 = 1-3y+3y^2-y^3 $$

Multiply $ \color{blue}{y} $ by $ \left( 1-2y+y^2\right) $ $$ \color{blue}{y} \cdot \left( 1-2y+y^2\right) = y-2y^2+y^3 $$

Add $ \dfrac{1-3y+3y^2-y^3}{3} $ and $ \dfrac{y-2y^2+y^3}{6} $ to get $ \dfrac{ \color{purple}{ -y^3+4y^2-5y+2 } }{ 6 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 2 }$.

$$ \begin{aligned} \frac{1-3y+3y^2-y^3}{3} + \frac{y-2y^2+y^3}{6} & = \frac{ \left( 1-3y+3y^2-y^3 \right) \cdot \color{blue}{ 2 }}{ 3 \cdot \color{blue}{ 2 }} + \frac{ y-2y^2+y^3 }{ 6 } = \\[1ex] &=\frac{ \color{purple}{ 2-6y+6y^2-2y^3 } }{ 6 } + \frac{ \color{purple}{ y-2y^2+y^3 } }{ 6 }=\frac{ \color{purple}{ -y^3+4y^2-5y+2 } }{ 6 } \end{aligned} $$