Divide $$$x^{3}$$$ by $$$x^{2} - 1$$$
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Your Input
Find $$$\frac{x^{3}}{x^{2} - 1}$$$ using long division.
Solution
Write the problem in the special format (missed terms are written with zero coefficients):
$$$\begin{array}{r|r}\hline\\x^{2}-1&x^{3}+0 x^{2}+0 x+0\end{array}$$$
Step 1
Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{x^{3}}{x^{2}} = x$$$.
Write down the calculated result in the upper part of the table.
Multiply it by the divisor: $$$x \left(x^{2}-1\right) = x^{3}- x$$$.
Subtract the dividend from the obtained result: $$$\left(x^{3}\right) - \left(x^{3}- x\right) = x$$$.
$$\begin{array}{r|rrrr:c}&{\color{Chartreuse}x}&&&&\\\hline\\{\color{Magenta}x^{2}}-1&{\color{Chartreuse}x^{3}}&+0 x^{2}&+0 x&+0&\frac{{\color{Chartreuse}x^{3}}}{{\color{Magenta}x^{2}}} = {\color{Chartreuse}x}\\&-\phantom{x^{3}}&&&&\\&x^{3}&+0 x^{2}&- x&&{\color{Chartreuse}x} \left(x^{2}-1\right) = x^{3}- x\\\hline\\&&&x&+0&\end{array}$$Since the degree of the remainder is less than the degree of the divisor, we are done.
The resulting table is shown once more:
$$\begin{array}{r|rrrr:c}&{\color{Chartreuse}x}&&&&\text{Hints}\\\hline\\{\color{Magenta}x^{2}}-1&{\color{Chartreuse}x^{3}}&+0 x^{2}&+0 x&+0&\frac{{\color{Chartreuse}x^{3}}}{{\color{Magenta}x^{2}}} = {\color{Chartreuse}x}\\&-\phantom{x^{3}}&&&&\\&x^{3}&+0 x^{2}&- x&&{\color{Chartreuse}x} \left(x^{2}-1\right) = x^{3}- x\\\hline\\&&&x&+0&\end{array}$$Therefore, $$$\frac{x^{3}}{x^{2} - 1} = x + \frac{x}{x^{2} - 1}$$$.
Answer
$$$\frac{x^{3}}{x^{2} - 1} = x + \frac{x}{x^{2} - 1}$$$A