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

The solution to expand ((4x^4-1)^2)/(4x^2) is 4x^6-2x^2+1/(4x^2)

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

expand

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

4 x^{6} - 2 x^{2} + \frac{1}{4 x ^{2}}

Solution steps

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

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

= \frac{16 x ^{8} - 8 x ^{4} + 1}{4 x ^{2}}

Apply the fraction rule:

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

\frac{16 x ^{8} - 8 x ^{4} + 1}{4 x ^{2}} = \frac{16 x ^{8}}{4 x ^{2}} - \frac{8 x ^{4}}{4 x ^{2}} + \frac{1}{4 x ^{2}}

= \frac{16 x ^{8}}{4 x ^{2}} - \frac{8 x ^{4}}{4 x ^{2}} + \frac{1}{4 x ^{2}}

Cancel

\frac{16 x ^{8}}{4 x ^{2}} : 4 x^{6}

= 4 x^{6} - \frac{8 x ^{4}}{4 x ^{2}} + \frac{1}{4 x ^{2}}

Cancel

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

= 4 x^{6} - 2 x^{2} + \frac{1}{4 x ^{2}}

(\frac{1}{3}) 5

5 (1 - 2 ln (x))

(a^{2} + 3 a + 2)^{6}

(x^{2} + 9) y^{'} (x) + x y

\frac{2}{( s ^{2} + 4 ) ( s ^{2} + 9 )}

What is expand ((4x^4-1)^2)/(4x^2) ?
The solution to expand ((4x^4-1)^2)/(4x^2) is 4x^6-2x^2+1/(4x^2)