What is expand-(4(9y^2+4x^2))/(81y^3) ?

The solution to expand-(4(9y^2+4x^2))/(81y^3) is -4/(9y)-(16x^2)/(81y^3)

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expand-(4(9y^2+4x^2))/(81y^3)

expand

- \frac{4 ( 9 y ^{2} + 4 x ^{2} )}{81 y ^{3}}

- \frac{4}{9 y} - \frac{16 x ^{2}}{81 y ^{3}}

Solution steps

- \frac{4 ( 9 y ^{2} + 4 x ^{2} )}{81 y ^{3}}

Expand

4 (9 y^{2} + 4 x^{2}) : 36 y^{2} + 16 x^{2}

= - \frac{36 y ^{2} + 16 x ^{2}}{81 y ^{3}}

Apply the fraction rule:

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

\frac{36 y ^{2} + 16 x ^{2}}{81 y ^{3}} = - (\frac{36 y ^{2}}{81 y ^{3}}) - (\frac{16 x ^{2}}{81 y ^{3}})

= - (\frac{36 y ^{2}}{81 y ^{3}}) - (\frac{16 x ^{2}}{81 y ^{3}})

Remove parentheses:

(a) = a

= - \frac{36 y ^{2}}{81 y ^{3}} - \frac{16 x ^{2}}{81 y ^{3}}

Cancel

\frac{36 y ^{2}}{81 y ^{3}} : \frac{4}{9 y}

= - \frac{4}{9 y} - \frac{16 x ^{2}}{81 y ^{3}}

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

\frac{s + 1}{( s ^{2} + 2 s + 10 ) ^{2}}

f^{'} (1) (x - 1)

\frac{( n + 8 ) ( n + 1 )}{n ( n - 10 )}

\frac{2 - pt}{2 ( 2 - p )}

What is expand-(4(9y^2+4x^2))/(81y^3) ?
The solution to expand-(4(9y^2+4x^2))/(81y^3) is -4/(9y)-(16x^2)/(81y^3)