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

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

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expand 1+(((4x^4-1)^2)/(16x^4))

expand

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

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

Solution steps

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

Remove parentheses:

(a) = a

= 1 + \frac{( 4 x ^{4} - 1 ) ^{2}}{16 x ^{4}}

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

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

Apply the fraction rule:

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

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

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

Cancel

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

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

Cancel

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

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

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

\frac{k ( 5 + 3 t )}{1 + b t}

\frac{s}{( s - 3 ) ^{2}}

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

- 2 (e^{2 x} C + 2 e^{2 x} (x C + d) + A)

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