What is expand ((y^4+1)^2)/(4y^4) ?

The solution to expand ((y^4+1)^2)/(4y^4) is (y^4)/4+1/2+1/(4y^4)

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expand ((y^4+1)^2)/(4y^4)

expand

\frac{( y ^{4} + 1 ) ^{2}}{4 y ^{4}}

\frac{y ^{4}}{4} + \frac{1}{2} + \frac{1}{4 y ^{4}}

Solution steps

\frac{( y ^{4} + 1 ) ^{2}}{4 y ^{4}}

(y^{4} + 1)^{2} = y^{8} + 2 y^{4} + 1

= \frac{y ^{8} + 2 y ^{4} + 1}{4 y ^{4}}

Apply the fraction rule:

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

\frac{y ^{8} + 2 y ^{4} + 1}{4 y ^{4}} = \frac{y ^{8}}{4 y ^{4}} + \frac{2 y ^{4}}{4 y ^{4}} + \frac{1}{4 y ^{4}}

= \frac{y ^{8}}{4 y ^{4}} + \frac{2 y ^{4}}{4 y ^{4}} + \frac{1}{4 y ^{4}}

Cancel

\frac{y ^{8}}{4 y ^{4}} : \frac{y ^{4}}{4}

= \frac{y ^{4}}{4} + \frac{2 y ^{4}}{4 y ^{4}} + \frac{1}{4 y ^{4}}

Cancel

\frac{2 y ^{4}}{4 y ^{4}} : \frac{1}{2}

= \frac{y ^{4}}{4} + \frac{1}{2} + \frac{1}{4 y ^{4}}

x - ((\frac{1}{x})^{\frac{1}{2}})^{60}

- e^{x} (x - 1)

(x + y)^{2} + (u + v)^{2}

\frac{( x + 1 ) ^{2} ( x - 5 )}{( x - 1 ) ^{3}}

\frac{2 e ^{- 2 s} ( 2 s ^{2} + 2 s + 1 )}{s ^{3}}

What is expand ((y^4+1)^2)/(4y^4) ?
The solution to expand ((y^4+1)^2)/(4y^4) is (y^4)/4+1/2+1/(4y^4)