What is expand ((a^2+9)(a^2-3)3)/(4a^2) ?

The solution to expand ((a^2+9)(a^2-3)3)/(4a^2) is (3a^2)/4+9/2-(81)/(4a^2)

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expand ((a^2+9)(a^2-3)3)/(4a^2)

expand

\frac{( a ^{2} + 9 ) ( a ^{2} - 3 ) 3}{4 a ^{2}}

\frac{3 a ^{2}}{4} + \frac{9}{2} - \frac{81}{4 a ^{2}}

Solution steps

\frac{( a ^{2} + 9 ) ( a ^{2} - 3 ) \cdot 3}{4 a ^{2}}

Expand

(a^{2} + 9) (a^{2} - 3) \cdot 3 : 3 a^{4} + 18 a^{2} - 81

= \frac{3 a ^{4} + 18 a ^{2} - 81}{4 a ^{2}}

Apply the fraction rule:

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

\frac{3 a ^{4} + 18 a ^{2} - 81}{4 a ^{2}} = \frac{3 a ^{4}}{4 a ^{2}} + \frac{18 a ^{2}}{4 a ^{2}} - \frac{81}{4 a ^{2}}

= \frac{3 a ^{4}}{4 a ^{2}} + \frac{18 a ^{2}}{4 a ^{2}} - \frac{81}{4 a ^{2}}

Cancel

\frac{3 a ^{4}}{4 a ^{2}} : \frac{3 a ^{2}}{4}

= \frac{3 a ^{2}}{4} + \frac{18 a ^{2}}{4 a ^{2}} - \frac{81}{4 a ^{2}}

Cancel

\frac{18 a ^{2}}{4 a ^{2}} : \frac{9}{2}

= \frac{3 a ^{2}}{4} + \frac{9}{2} - \frac{81}{4 a ^{2}}

\frac{1}{s ( s ^{2} + A ^{2} )}

4 + f (2 t - 7) H (t - 2)

\frac{2}{3} n - \frac{1}{4}

\frac{x ( x - 1 )}{( x - 2 ) ( x - 4 )}

- 6 (- 6 e^{(- 3 x^{2})} (x^{2}) + e^{(- 3 x^{2})})

What is expand ((a^2+9)(a^2-3)3)/(4a^2) ?
The solution to expand ((a^2+9)(a^2-3)3)/(4a^2) is (3a^2)/4+9/2-(81)/(4a^2)