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

The solution to expand ((1+x)(1+2x)(1+3x))/x is 1/x+6+11x+6x^2

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

expand

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

\frac{1}{x} + 6 + 11 x + 6 x^{2}

Solution steps

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

Expand

(1 + x) (1 + 2 x) (1 + 3 x) : 1 + 6 x + 11 x^{2} + 6 x^{3}

= \frac{1 + 6 x + 11 x ^{2} + 6 x ^{3}}{x}

Apply the fraction rule:

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

\frac{1 + 6 x + 11 x ^{2} + 6 x ^{3}}{x} = \frac{1}{x} + \frac{6 x}{x} + \frac{11 x ^{2}}{x} + \frac{6 x ^{3}}{x}

= \frac{1}{x} + \frac{6 x}{x} + \frac{11 x ^{2}}{x} + \frac{6 x ^{3}}{x}

Cancel

\frac{6 x}{x} : 6

= \frac{1}{x} + 6 + \frac{11 x ^{2}}{x} + \frac{6 x ^{3}}{x}

Cancel

\frac{11 x ^{2}}{x} : 11 x

= \frac{1}{x} + 6 + 11 x + \frac{6 x ^{3}}{x}

Cancel

\frac{6 x ^{3}}{x} : 6 x^{2}

= \frac{1}{x} + 6 + 11 x + 6 x^{2}

π^{8} (19 - 2 - y^{2})^{2}

\frac{- 2 x ^{5} + 2 x}{( x ^{4} + 1 ) ^{2}}

36 (π - 2)

\frac{( x ^{2} - 2 x - 15 ) ( x + 3 )}{( - x ^{2} - 1 ) ( x - 2 )}

l im x \Rightarrow - 2 (x^{2} - x - 6)

What is expand ((1+x)(1+2x)(1+3x))/x ?
The solution to expand ((1+x)(1+2x)(1+3x))/x is 1/x+6+11x+6x^2