What is expand ((4y^4-1)^2)/(16y^4) ?

The solution to expand ((4y^4-1)^2)/(16y^4) is y^4-1/2+1/(16y^4)

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expand ((4y^4-1)^2)/(16y^4)

expand

\frac{( 4 y ^{4} - 1 ) ^{2}}{16 y ^{4}}

y^{4} - \frac{1}{2} + \frac{1}{16 y ^{4}}

Solution steps

\frac{( 4 y ^{4} - 1 ) ^{2}}{16 y ^{4}}

(4 y^{4} - 1)^{2} = 16 y^{8} - 8 y^{4} + 1

= \frac{16 y ^{8} - 8 y ^{4} + 1}{16 y ^{4}}

Apply the fraction rule:

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

\frac{16 y ^{8} - 8 y ^{4} + 1}{16 y ^{4}} = \frac{16 y ^{8}}{16 y ^{4}} - \frac{8 y ^{4}}{16 y ^{4}} + \frac{1}{16 y ^{4}}

= \frac{16 y ^{8}}{16 y ^{4}} - \frac{8 y ^{4}}{16 y ^{4}} + \frac{1}{16 y ^{4}}

Cancel

\frac{16 y ^{8}}{16 y ^{4}} : y^{4}

= y^{4} - \frac{8 y ^{4}}{16 y ^{4}} + \frac{1}{16 y ^{4}}

Cancel

\frac{8 y ^{4}}{16 y ^{4}} : \frac{1}{2}

= y^{4} - \frac{1}{2} + \frac{1}{16 y ^{4}}

\frac{1 - 2 ( a + h ) - 1 - 2 a}{h}

\frac{s + 3}{( s + 1 ) ( s + 2 ) ^{2} ( s + 4 )}

f (- 8 sec (x) tan (x) + 9 sec^{2} (x)) d x

\frac{( 1 + 2 x ) ^{3}}{x ^{4}}

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

What is expand ((4y^4-1)^2)/(16y^4) ?
The solution to expand ((4y^4-1)^2)/(16y^4) is y^4-1/2+1/(16y^4)