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

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

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

expand

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

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

Solution steps

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

Expand

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

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

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

\frac{x}{x} = 1

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

Cancel

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

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

Cancel

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

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

2 x + 2 x - 2 x

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

- \frac{( x - 4 ) ( x - 16 )}{2}

(4 - 2.6) d i v i d e [,] (8.1 - 2.6)

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

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