Divide $$$x^{3} - 1$$$ by $$$x - 2$$$

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

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

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: $$$x^{2} \left(x-2\right) = x^{3}- 2 x^{2}$$$.

Subtract the dividend from the obtained result: $$$\left(x^{3}-1\right) - \left(x^{3}- 2 x^{2}\right) = 2 x^{2}-1$$$.

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

Step 2

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: $$$\left(2 x^{2}-1\right) - \left(2 x^{2}- 4 x\right) = 4 x-1$$$.

$$\begin{array}{r|rrrr:c}&x^{2}&{\color{DarkCyan}+2 x}&&&\\\hline\\{\color{Magenta}x}-2&x^{3}&+0 x^{2}&+0 x&-1&\\&-\phantom{x^{3}}&&&&\\&x^{3}&- 2 x^{2}&&&\\\hline\\&&{\color{DarkCyan}2 x^{2}}&+0 x&-1&\frac{{\color{DarkCyan}2 x^{2}}}{{\color{Magenta}x}} = {\color{DarkCyan}2 x}\\&&-\phantom{2 x^{2}}&&&\\&&2 x^{2}&- 4 x&&{\color{DarkCyan}2 x} \left(x-2\right) = 2 x^{2}- 4 x\\\hline\\&&&4 x&-1&\end{array}$$

Step 3

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: $$$\left(4 x-1\right) - \left(4 x-8\right) = 7$$$.

$$\begin{array}{r|rrrr:c}&x^{2}&+2 x&{\color{Red}+4}&&\\\hline\\{\color{Magenta}x}-2&x^{3}&+0 x^{2}&+0 x&-1&\\&-\phantom{x^{3}}&&&&\\&x^{3}&- 2 x^{2}&&&\\\hline\\&&2 x^{2}&+0 x&-1&\\&&-\phantom{2 x^{2}}&&&\\&&2 x^{2}&- 4 x&&\\\hline\\&&&{\color{Red}4 x}&-1&\frac{{\color{Red}4 x}}{{\color{Magenta}x}} = {\color{Red}4}\\&&&-\phantom{4 x}&&\\&&&4 x&-8&{\color{Red}4} \left(x-2\right) = 4 x-8\\\hline\\&&&&7&\end{array}$$

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

The resulting table is shown once more:

$$\begin{array}{r|rrrr:c}&{\color{Green}x^{2}}&{\color{DarkCyan}+2 x}&{\color{Red}+4}&&\text{Hints}\\\hline\\{\color{Magenta}x}-2&{\color{Green}x^{3}}&+0 x^{2}&+0 x&-1&\frac{{\color{Green}x^{3}}}{{\color{Magenta}x}} = {\color{Green}x^{2}}\\&-\phantom{x^{3}}&&&&\\&x^{3}&- 2 x^{2}&&&{\color{Green}x^{2}} \left(x-2\right) = x^{3}- 2 x^{2}\\\hline\\&&{\color{DarkCyan}2 x^{2}}&+0 x&-1&\frac{{\color{DarkCyan}2 x^{2}}}{{\color{Magenta}x}} = {\color{DarkCyan}2 x}\\&&-\phantom{2 x^{2}}&&&\\&&2 x^{2}&- 4 x&&{\color{DarkCyan}2 x} \left(x-2\right) = 2 x^{2}- 4 x\\\hline\\&&&{\color{Red}4 x}&-1&\frac{{\color{Red}4 x}}{{\color{Magenta}x}} = {\color{Red}4}\\&&&-\phantom{4 x}&&\\&&&4 x&-8&{\color{Red}4} \left(x-2\right) = 4 x-8\\\hline\\&&&&7&\end{array}$$

Therefore, $$$\frac{x^{3} - 1}{x - 2} = \left(x^{2} + 2 x + 4\right) + \frac{7}{x - 2}$$$.