What is expand ((2-t^5)^3)/(8t^6) ?

The solution to expand ((2-t^5)^3)/(8t^6) is 1/(t^6)-3/(2t)+(3t^4)/4-(t^9)/8

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expand ((2-t^5)^3)/(8t^6)

expand

\frac{( 2 - t ^{5} ) ^{3}}{8 t ^{6}}

\frac{1}{t ^{6}} - \frac{3}{2 t} + \frac{3 t ^{4}}{4} - \frac{t ^{9}}{8}

Solution steps

\frac{( 2 - t ^{5} ) ^{3}}{8 t ^{6}}

(2 - t^{5})^{3} = 8 - 12 t^{5} + 6 t^{10} - t^{15}

= \frac{8 - 12 t ^{5} + 6 t ^{10} - t ^{15}}{8 t ^{6}}

Apply the fraction rule:

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

\frac{8 - 12 t ^{5} + 6 t ^{10} - t ^{15}}{8 t ^{6}} = \frac{8}{8 t ^{6}} - \frac{12 t ^{5}}{8 t ^{6}} + \frac{6 t ^{10}}{8 t ^{6}} - \frac{t ^{15}}{8 t ^{6}}

= \frac{8}{8 t ^{6}} - \frac{12 t ^{5}}{8 t ^{6}} + \frac{6 t ^{10}}{8 t ^{6}} - \frac{t ^{15}}{8 t ^{6}}

\frac{8}{8 t ^{6}} = \frac{1}{t ^{6}}

\frac{12 t ^{5}}{8 t ^{6}} = \frac{3}{2 t}

\frac{6 t ^{10}}{8 t ^{6}} = \frac{3 t ^{4}}{4}

\frac{t ^{15}}{8 t ^{6}} = \frac{t ^{9}}{8}

= \frac{1}{t ^{6}} - \frac{3}{2 t} + \frac{3 t ^{4}}{4} - \frac{t ^{9}}{8}

\frac{x ( x - 3 )}{( x - 1 ) ( x - 5 )}

(y^{1} = 2, - x^{2}) (y_{2} = - x)

\frac{( 3 + X ) ( 3 - X )}{X ^{2}}

\frac{d}{d t} ((t) k (t - 70))

483 \cdot \frac{1}{6}

What is expand ((2-t^5)^3)/(8t^6) ?
The solution to expand ((2-t^5)^3)/(8t^6) is 1/(t^6)-3/(2t)+(3t^4)/4-(t^9)/8