Divide $$$x^{3} \left(x - 1\right)$$$ by $$$x - 2$$$

Rewrite the dividend: $$$x^{3} \left(x - 1\right) = x^{4} - x^{3}$$$.

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

$$$\begin{array}{r|r}\hline\\x-2&x^{4}- x^{3}+0 x^{2}+0 x+0\end{array}$$$

Step 1

Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{x^{4}}{x} = x^{3}$$$.

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

Multiply it by the divisor: $$$x^{3} \left(x-2\right) = x^{4}- 2 x^{3}$$$.

Subtract the dividend from the obtained result: $$$\left(x^{4}- x^{3}\right) - \left(x^{4}- 2 x^{3}\right) = x^{3}$$$.

$$\begin{array}{r|rrrrr:c}&{\color{DarkMagenta}x^{3}}&&&&&\\\hline\\{\color{Magenta}x}-2&{\color{DarkMagenta}x^{4}}&- x^{3}&+0 x^{2}&+0 x&+0&\frac{{\color{DarkMagenta}x^{4}}}{{\color{Magenta}x}} = {\color{DarkMagenta}x^{3}}\\&-\phantom{x^{4}}&&&&&\\&x^{4}&- 2 x^{3}&&&&{\color{DarkMagenta}x^{3}} \left(x-2\right) = x^{4}- 2 x^{3}\\\hline\\&&x^{3}&+0 x^{2}&+0 x&+0&\end{array}$$

Step 2

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

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

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

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

$$\begin{array}{r|rrrrr:c}&x^{3}&{\color{Chartreuse}+x^{2}}&&&&\\\hline\\{\color{Magenta}x}-2&x^{4}&- x^{3}&+0 x^{2}&+0 x&+0&\\&-\phantom{x^{4}}&&&&&\\&x^{4}&- 2 x^{3}&&&&\\\hline\\&&{\color{Chartreuse}x^{3}}&+0 x^{2}&+0 x&+0&\frac{{\color{Chartreuse}x^{3}}}{{\color{Magenta}x}} = {\color{Chartreuse}x^{2}}\\&&-\phantom{x^{3}}&&&&\\&&x^{3}&- 2 x^{2}&&&{\color{Chartreuse}x^{2}} \left(x-2\right) = x^{3}- 2 x^{2}\\\hline\\&&&2 x^{2}&+0 x&+0&\end{array}$$

Step 3

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

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

Multiply it by the divisor: $$$2 x \left(x-2\right) = 2 x^{2}- 4 x$$$.

Subtract the remainder from the obtained result: $$$\left(2 x^{2}\right) - \left(2 x^{2}- 4 x\right) = 4 x$$$.

$$\begin{array}{r|rrrrr:c}&x^{3}&+x^{2}&{\color{OrangeRed}+2 x}&&&\\\hline\\{\color{Magenta}x}-2&x^{4}&- x^{3}&+0 x^{2}&+0 x&+0&\\&-\phantom{x^{4}}&&&&&\\&x^{4}&- 2 x^{3}&&&&\\\hline\\&&x^{3}&+0 x^{2}&+0 x&+0&\\&&-\phantom{x^{3}}&&&&\\&&x^{3}&- 2 x^{2}&&&\\\hline\\&&&{\color{OrangeRed}2 x^{2}}&+0 x&+0&\frac{{\color{OrangeRed}2 x^{2}}}{{\color{Magenta}x}} = {\color{OrangeRed}2 x}\\&&&-\phantom{2 x^{2}}&&&\\&&&2 x^{2}&- 4 x&&{\color{OrangeRed}2 x} \left(x-2\right) = 2 x^{2}- 4 x\\\hline\\&&&&4 x&+0&\end{array}$$

Step 4

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{4 x}{x} = 4$$$.

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

Multiply it by the divisor: $$$4 \left(x-2\right) = 4 x-8$$$.

Subtract the remainder from the obtained result: $$$\left(4 x\right) - \left(4 x-8\right) = 8$$$.

$$\begin{array}{r|rrrrr:c}&x^{3}&+x^{2}&+2 x&{\color{Chocolate}+4}&&\\\hline\\{\color{Magenta}x}-2&x^{4}&- x^{3}&+0 x^{2}&+0 x&+0&\\&-\phantom{x^{4}}&&&&&\\&x^{4}&- 2 x^{3}&&&&\\\hline\\&&x^{3}&+0 x^{2}&+0 x&+0&\\&&-\phantom{x^{3}}&&&&\\&&x^{3}&- 2 x^{2}&&&\\\hline\\&&&2 x^{2}&+0 x&+0&\\&&&-\phantom{2 x^{2}}&&&\\&&&2 x^{2}&- 4 x&&\\\hline\\&&&&{\color{Chocolate}4 x}&+0&\frac{{\color{Chocolate}4 x}}{{\color{Magenta}x}} = {\color{Chocolate}4}\\&&&&-\phantom{4 x}&&\\&&&&4 x&-8&{\color{Chocolate}4} \left(x-2\right) = 4 x-8\\\hline\\&&&&&8&\end{array}$$

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

The resulting table is shown once more:

$$\begin{array}{r|rrrrr:c}&{\color{DarkMagenta}x^{3}}&{\color{Chartreuse}+x^{2}}&{\color{OrangeRed}+2 x}&{\color{Chocolate}+4}&&\text{Hints}\\\hline\\{\color{Magenta}x}-2&{\color{DarkMagenta}x^{4}}&- x^{3}&+0 x^{2}&+0 x&+0&\frac{{\color{DarkMagenta}x^{4}}}{{\color{Magenta}x}} = {\color{DarkMagenta}x^{3}}\\&-\phantom{x^{4}}&&&&&\\&x^{4}&- 2 x^{3}&&&&{\color{DarkMagenta}x^{3}} \left(x-2\right) = x^{4}- 2 x^{3}\\\hline\\&&{\color{Chartreuse}x^{3}}&+0 x^{2}&+0 x&+0&\frac{{\color{Chartreuse}x^{3}}}{{\color{Magenta}x}} = {\color{Chartreuse}x^{2}}\\&&-\phantom{x^{3}}&&&&\\&&x^{3}&- 2 x^{2}&&&{\color{Chartreuse}x^{2}} \left(x-2\right) = x^{3}- 2 x^{2}\\\hline\\&&&{\color{OrangeRed}2 x^{2}}&+0 x&+0&\frac{{\color{OrangeRed}2 x^{2}}}{{\color{Magenta}x}} = {\color{OrangeRed}2 x}\\&&&-\phantom{2 x^{2}}&&&\\&&&2 x^{2}&- 4 x&&{\color{OrangeRed}2 x} \left(x-2\right) = 2 x^{2}- 4 x\\\hline\\&&&&{\color{Chocolate}4 x}&+0&\frac{{\color{Chocolate}4 x}}{{\color{Magenta}x}} = {\color{Chocolate}4}\\&&&&-\phantom{4 x}&&\\&&&&4 x&-8&{\color{Chocolate}4} \left(x-2\right) = 4 x-8\\\hline\\&&&&&8&\end{array}$$

Therefore, $$$\frac{x^{3} \left(x - 1\right)}{x - 2} = \left(x^{3} + x^{2} + 2 x + 4\right) + \frac{8}{x - 2}$$$.