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

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

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

expand

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

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

Solution steps

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

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

= \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}

= 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}

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

(a + b)^{2}

(2 + 4 - x)^{2}

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

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

(3 x^{2} + 7 x^{3}) e^{7 x}

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