Divide $$$y^{3}$$$ by $$$1 - y$$$

Write the problem in the special format (missed terms are written with zero coefficients):

$$$\begin{array}{r|r}\hline\\- y+1&y^{3}+0 y^{2}+0 y+0\end{array}$$$

Step 1

Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{y^{3}}{- y} = - y^{2}$$$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $$$- y^{2} \left(- y+1\right) = y^{3}- y^{2}$$$.

Subtract the dividend from the obtained result: $$$\left(y^{3}\right) - \left(y^{3}- y^{2}\right) = y^{2}$$$.

$$\begin{array}{r|rrrr:c}&{\color{Violet}- y^{2}}&&&&\\\hline\\{\color{Magenta}- y}+1&{\color{Violet}y^{3}}&+0 y^{2}&+0 y&+0&\frac{{\color{Violet}y^{3}}}{{\color{Magenta}- y}} = {\color{Violet}- y^{2}}\\&-\phantom{y^{3}}&&&&\\&y^{3}&- y^{2}&&&{\color{Violet}- y^{2}} \left(- y+1\right) = y^{3}- y^{2}\\\hline\\&&y^{2}&+0 y&+0&\end{array}$$

Step 2

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{y^{2}}{- y} = - y$$$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $$$- y \left(- y+1\right) = y^{2}- y$$$.

Subtract the remainder from the obtained result: $$$\left(y^{2}\right) - \left(y^{2}- y\right) = y$$$.

$$\begin{array}{r|rrrr:c}&- y^{2}&{\color{Fuchsia}- y}&&&\\\hline\\{\color{Magenta}- y}+1&y^{3}&+0 y^{2}&+0 y&+0&\\&-\phantom{y^{3}}&&&&\\&y^{3}&- y^{2}&&&\\\hline\\&&{\color{Fuchsia}y^{2}}&+0 y&+0&\frac{{\color{Fuchsia}y^{2}}}{{\color{Magenta}- y}} = {\color{Fuchsia}- y}\\&&-\phantom{y^{2}}&&&\\&&y^{2}&- y&&{\color{Fuchsia}- y} \left(- y+1\right) = y^{2}- y\\\hline\\&&&y&+0&\end{array}$$

Step 3

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{y}{- y} = -1$$$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $$$- \left(- y+1\right) = y-1$$$.

Subtract the remainder from the obtained result: $$$\left(y\right) - \left(y-1\right) = 1$$$.

$$\begin{array}{r|rrrr:c}&- y^{2}&- y&{\color{Brown}-1}&&\\\hline\\{\color{Magenta}- y}+1&y^{3}&+0 y^{2}&+0 y&+0&\\&-\phantom{y^{3}}&&&&\\&y^{3}&- y^{2}&&&\\\hline\\&&y^{2}&+0 y&+0&\\&&-\phantom{y^{2}}&&&\\&&y^{2}&- y&&\\\hline\\&&&{\color{Brown}y}&+0&\frac{{\color{Brown}y}}{{\color{Magenta}- y}} = {\color{Brown}-1}\\&&&-\phantom{y}&&\\&&&y&-1&{\color{Brown}-1} \left(- y+1\right) = y-1\\\hline\\&&&&1&\end{array}$$

Since the degree of the remainder is less than the degree of the divisor, we are done.

The resulting table is shown once more:

$$\begin{array}{r|rrrr:c}&{\color{Violet}- y^{2}}&{\color{Fuchsia}- y}&{\color{Brown}-1}&&\text{Hints}\\\hline\\{\color{Magenta}- y}+1&{\color{Violet}y^{3}}&+0 y^{2}&+0 y&+0&\frac{{\color{Violet}y^{3}}}{{\color{Magenta}- y}} = {\color{Violet}- y^{2}}\\&-\phantom{y^{3}}&&&&\\&y^{3}&- y^{2}&&&{\color{Violet}- y^{2}} \left(- y+1\right) = y^{3}- y^{2}\\\hline\\&&{\color{Fuchsia}y^{2}}&+0 y&+0&\frac{{\color{Fuchsia}y^{2}}}{{\color{Magenta}- y}} = {\color{Fuchsia}- y}\\&&-\phantom{y^{2}}&&&\\&&y^{2}&- y&&{\color{Fuchsia}- y} \left(- y+1\right) = y^{2}- y\\\hline\\&&&{\color{Brown}y}&+0&\frac{{\color{Brown}y}}{{\color{Magenta}- y}} = {\color{Brown}-1}\\&&&-\phantom{y}&&\\&&&y&-1&{\color{Brown}-1} \left(- y+1\right) = y-1\\\hline\\&&&&1&\end{array}$$

Therefore, $$$\frac{y^{3}}{1 - y} = \left(- y^{2} - y - 1\right) + \frac{1}{1 - y}$$$.