What is expand ((x^2-3)^2)/(x^5) ?

The solution to expand ((x^2-3)^2)/(x^5) is 1/x-6/(x^3)+9/(x^5)

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expand ((x^2-3)^2)/(x^5)

expand

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

\frac{1}{x} - \frac{6}{x ^{3}} + \frac{9}{x ^{5}}

Solution steps

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

(x^{2} - 3)^{2} = x^{4} - 6 x^{2} + 9

= \frac{x ^{4} - 6 x ^{2} + 9}{x ^{5}}

Apply the fraction rule:

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

\frac{x ^{4} - 6 x ^{2} + 9}{x ^{5}} = \frac{x ^{4}}{x ^{5}} - \frac{6 x ^{2}}{x ^{5}} + \frac{9}{x ^{5}}

= \frac{x ^{4}}{x ^{5}} - \frac{6 x ^{2}}{x ^{5}} + \frac{9}{x ^{5}}

Cancel

\frac{x ^{4}}{x ^{5}} : \frac{1}{x}

= \frac{1}{x} - \frac{6 x ^{2}}{x ^{5}} + \frac{9}{x ^{5}}

Cancel

\frac{6 x ^{2}}{x ^{5}} : \frac{6}{x ^{3}}

= \frac{1}{x} - \frac{6}{x ^{3}} + \frac{9}{x ^{5}}

\frac{1}{( 1 - a x ) ( 1 - b x ) ( 1 - c x )}

(3 d + 8 e)^{3}

- 3 x (3 x + 2 x ln (x))

(\frac{x}{4 y ^{'} ( x )} + y)^{2}

p u n t os (2, - 1) y (\frac{1 , - 9}{7})

What is expand ((x^2-3)^2)/(x^5) ?
The solution to expand ((x^2-3)^2)/(x^5) is 1/x-6/(x^3)+9/(x^5)