What is expand ((x^2+1)(x^3-1))/x ?

The solution to expand ((x^2+1)(x^3-1))/x is x^4-x+x^2-1/x

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expand ((x^2+1)(x^3-1))/x

expand

\frac{( x ^{2} + 1 ) ( x ^{3} - 1 )}{x}

x^{4} - x + x^{2} - \frac{1}{x}

Solution steps

\frac{( x ^{2} + 1 ) ( x ^{3} - 1 )}{x}

Expand

(x^{2} + 1) (x^{3} - 1) : x^{5} - x^{2} + x^{3} - 1

= \frac{x ^{5} - x ^{2} + x ^{3} - 1}{x}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{x ^{5} - x ^{2} + x ^{3} - 1}{x} = \frac{x ^{5}}{x} - \frac{x ^{2}}{x} + \frac{x ^{3}}{x} - \frac{1}{x}

= \frac{x ^{5}}{x} - \frac{x ^{2}}{x} + \frac{x ^{3}}{x} - \frac{1}{x}

Cancel

\frac{x ^{5}}{x} : x^{4}

= x^{4} - \frac{x ^{2}}{x} + \frac{x ^{3}}{x} - \frac{1}{x}

Cancel

\frac{x ^{2}}{x} : x

= x^{4} - x + \frac{x ^{3}}{x} - \frac{1}{x}

Cancel

\frac{x ^{3}}{x} : x^{2}

= x^{4} - x + x^{2} - \frac{1}{x}

4 (2 sin (x - 2 t) + 2 e^{x + 2 t})

\frac{( ( x + 1 ) ( x + 2 ) ( 2 x + 3 ) )}{6}

\frac{6 x - 42}{( x - 4 ) ( x - 6 )}

cos (π (0.5 + h))

- \frac{5}{( x + 2 ) ( x - 3 )}

What is expand ((x^2+1)(x^3-1))/x ?
The solution to expand ((x^2+1)(x^3-1))/x is x^4-x+x^2-1/x