What is expand (x(x^2-4)^2)/(12y) ?

The solution to expand (x(x^2-4)^2)/(12y) is (x^5)/(12y)-(2x^3)/(3y)+(4x)/(3y)

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expand (x(x^2-4)^2)/(12y)

expand

\frac{x ( x ^{2} - 4 ) ^{2}}{12 y}

\frac{x ^{5}}{12 y} - \frac{2 x ^{3}}{3 y} + \frac{4 x}{3 y}

Solution steps

\frac{x ( x ^{2} - 4 ) ^{2}}{12 y}

(x^{2} - 4)^{2} = x^{4} - 8 x^{2} + 16

= \frac{x ( x ^{4} - 8 x ^{2} + 16 )}{12 y}

Expand

x (x^{4} - 8 x^{2} + 16) : x^{5} - 8 x^{3} + 16 x

= \frac{x ^{5} - 8 x ^{3} + 16 x}{12 y}

Apply the fraction rule:

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

\frac{x ^{5} - 8 x ^{3} + 16 x}{12 y} = \frac{x ^{5}}{12 y} - \frac{8 x ^{3}}{12 y} + \frac{16 x}{12 y}

= \frac{x ^{5}}{12 y} - \frac{8 x ^{3}}{12 y} + \frac{16 x}{12 y}

Cancel

\frac{8 x ^{3}}{12 y} : \frac{2 x ^{3}}{3 y}

= \frac{x ^{5}}{12 y} - \frac{2 x ^{3}}{3 y} + \frac{16 x}{12 y}

Cancel

\frac{16 x}{12 y} : \frac{4 x}{3 y}

= \frac{x ^{5}}{12 y} - \frac{2 x ^{3}}{3 y} + \frac{4 x}{3 y}

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\frac{r \cdot ( 4 r ^{2} sin ^{2} ( θ ) - r ^{4} )}{2}

5 (7 - x = 31 - x)

(1 - \frac{4}{x}) e^{x}

What is expand (x(x^2-4)^2)/(12y) ?
The solution to expand (x(x^2-4)^2)/(12y) is (x^5)/(12y)-(2x^3)/(3y)+(4x)/(3y)