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

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

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

Step 1

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

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

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

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

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

Step 2

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

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

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

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

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

Step 3

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

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

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

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

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

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

The resulting table is shown once more:

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

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